In this video, we're going to be
solving whole collection of

trigonometric equations now be
cause it's the technique of

solving the equation and in
ensuring that we get enough

solutions, that's important and
not actually looking up the

angle. All of these are designed
around certain special angles,

so I'm just going to list at the
very beginning here the special

angles and their sines, cosines,
and tangents that are going to

form. The basis of what
we're doing.

So the special angles that we're
going to have a look at our

zero. 30 4560
and 90 there in degrees.

If we're thinking about radians,
then there's zero.

Pie by 6.

Pie by 4.

Pie by three.

And Π by 2.

Trig ratios we're going to be
looking at are the sign.

The cosine. On the tangent
of each of these.

Sign of 0 is 0.

The sign of 30 is 1/2.

Sign of 45 is one over Route 2.

The sign of 60 is Route 3 over 2
and the sign of 90 is one.

Cosine of 0 is one.

Cosine of 30 is Route 3 over 2
cosine of 45 is one over Route

2, the cosine of 60 is 1/2, and
the cosine of 90 is 0.

The town of 0 is 0 the
town of 30 is one over

Route 3 that Anna 45 is
110 of 60 is Route 3 and

the town of 90 degrees is
infinite, it's undefined.

It's these that we're going to
be looking at and working with.

Let's look at our first

equation then. We're going to
begin with some very simple

ones. So we take sign of
X is equal to nought .5. Now

invariably when we get an
equation we get a range of

values along with it.

So in this case will take X is
between North and 360. So what

we're looking for is all the
values of X.

Husain gives us N
.5.

Let's sketch a graph of
sine X over this range.

And sign looks like that
with 90.

180

270 and 360 and ranging between
one 4 sign 90 and minus

one for the sign of 270.

Sign of X is nought .5. So
we go there.

And there.

So there's our first angle, and
there's our second angle.

We know the first one is 30
degrees because sign of 30 is

1/2, so our first angle is 30
degrees. This curve is symmetric

and so because were 30 degrees
in from there, this one's got to

be 30 degrees back from there.

That would make it

150. There are no more answers
because within this range as we

go along this line.

It doesn't cross the curve at
any other points.

Let's have a look
at a cosine cause

of X is minus
nought .5 and the

range for this X
between North and 360.

So again, let's have a look at a
graph of the function.

Involved in the equation,
the cosine graph.

Looks like that. One and

minus one.
This is 90.

180

270 and then here
at the end, 360.

Minus 9.5.

Gain across there at minus
9.5 and down to their and

down to their.

Now the one thing we do know is
that the cause of 60 is plus N

.5, and so that's there. So we
know there is 60. Now again,

this curve is symmetric, so if
that one is 30 back that way

this one must be 30 further on.
So I'll first angle must be 120

degrees. This one's got to be in
a similar position as this bit

of the curve is again symmetric.

So that's 270 and we need
to come back 30 degrees, so

that's 240. Now we're going to
have a look at an example where

we've got what we call on
multiple angle. So instead of

just being cause of X or sign of
X, it's going to be something

like sign of 2X or cause of
three X. So let's begin with

sign of. 2X is equal
to Route 3 over 2

and again will take X
to be between North and

360.

Now we've got 2X here.

So if we've got 2X and X
is between Norton 360, then the

total range that we're going to
be looking at is not to 722.

X is going to come between 0
and 720, and the sign function

is periodic. It repeats itself
every 360 degrees, so I'm going

to need 2 copies of the sine

curve. As the first one going up
to 360 and now I need a second

copy there going on till.

720

OK, so sign 2 X equals root, 3
over 2, but we know that the

sign of 60 is Route 3 over 2. So
if we put in Route 3 over 2 it's

there, then it's going to be
these along here as well. So

what have we got? Well, the
first one here we know is 60.

This point we know is 180 so
that one's got to be the same

distance. Back in due to the
symmetry 120, so we do know that

2X will be 60 or 120, but we
also now we've got these other

points on here, so let's just
count on where we are. There's

the 1st loop of the sign
function, the first copy, its

periodic and repeats itself
again. So now we need to know

where are these well.

This is an exact copy of
that, so this must be 60

further on. In other words, at
420, and this must be another

120 further on. In other words,
at 480. So we've got two

more answers. And it's X
that we actually want, not

2X. So this is 3060.

210 and finally

240. Let's have a
look at that with a tangent

function. This time tan or three
X is equal to.

Minus one and will
take X to be

between North and 180.

So we draw a graph
of the tangent function.

So we go up.

We've got that there. That's 90.

This is 180 and this is 270
now. It's 3X. X is between

Norton 180, so 3X can be between
North and 3 * 180 which is

540. So I need to get copies
of this using the periodicity of

the tangent function right up to
540. So let's put in some more.

That's 360. On
there.

That's 450.

This one here will be
540 and that's as near

or as far as we
need to go. Tanner 3X

is minus one, so here's

minus one. And we go across here
picking off all the ones that we

need. So we've got one there.

There there. These are our
values, so 3X is equal 12.

Now we know that the angle
whose tangent is one is 45,

which is there. So again this
and this are symmetric bits of

curve, so this must be 45
further on. In other words 130.

5.

This one here has got to be
45 further on, so that will be

315. This one here has got
to be 45 further on, so that

will be 495, but it's X that
we want not 3X, so let's divide

throughout by three, so freezing
to that is 45 threes into that

is 105 and threes into that is
165. Those are our three answers

for that one 45 degrees.

105 degrees under

165. Let's take cause of
X over 2 this time. So

instead of multiplying by two or
by three, were now dividing by

two. Let's see what difference
this might make equals minus 1/2

and will take X to be
between North and 360. So let's

draw the graph.

All calls X between North
and 360, so there we've

got it 360 there.

180 there, we've got 90 and 270
there in their minus. 1/2 now

that's going to be.

Their cross and then these are
the ones that we are after.

So let's work with that. X over
2 is equal tool. Now where are

we? Well, we know that the angle
whose cosine is 1/2 is in fact

60 degrees, which is here 30 in
from there. So that must be 30

further on. In other words, 120
and this one must be 30 back. In

other words, 240. So now we
multiply it by.

Two, we get 240 and 480, but
of course this one is outside

the given range. The range is
not to 360, so we do not

need that answer, just want the

240. Now we've been working with
a range of North 360, or in one

case not to 180, so let's change
the range now so it's a

symmetric range in the Y axis,
so the range is now going to run

from minus 180 to plus 180

degrees. So we'll begin with
sign of X equals 1X is to

be between 180 degrees but
greater than minus 180 degrees.

Let's sketch the graph of sign
in that range. So we want to

complete copy of it. It's going
to look like that.

Now we know that the angle
who sign is one is 90

degrees and so we know
that's one there and that's

90 there and we can see that
there is only the one

solution it meets the curve
once and once only, so

that's 90 degrees.

Once and once only, that is
within the defined range. Let's

take another one.

So now we use a multiple
angle cause 2 X equals 1/2

and will take X to be
between minus 180 degrees and

plus 180 degrees. So let's
sketch the graph. Let's remember

that if X is between minus
180 and plus one 80, then

2X will be between minus 360.

And plus 360.

So what we need to do is use the
periodicity of the cosine

function to sketch it.

In the range. So there's
the knocked 360 bit and

then we want.

To minus 360. So I just label up
the points. Here is 90.

180

Two 7360 and then back
this way minus 90 -

180. Minus 270 and

minus 360. Now cause
2X is 1/2, so here's a half.

Membrane that this goes between
plus one and minus one and if we

draw a line across to see where
it meets the curve.

Then we can see it meets it in
four places. There, there there

and there we know that the angle
where it meets here is 60

degrees. So our first value is 2
X equals 60 degrees.

By symmetry, this one back here
has got to be minus 60.

What about this one here? Well,
again, symmetry says that we are

60 from here, so we've got to
be 60 back from there, so this

must be 300 and our symmetry of
the curve says that this one

must be minus 300, and so we
have X is 30 degrees minus 30

degrees, 150 degrees and minus

150 degrees. Working with
the tangent function tan, two

X equals Route 3 and
again will place X between

180 degrees and minus 180
degrees. We want to sketch

the function for tangent and
we want to be aware

that we've got 2X.

So since X is between minus 118
+ 182, X is got to be between

minus 360 and plus 360.

So if we take the bit between.

North And 360.

Which is that bit of the curve
we need a copy of that between

minus 360 and 0 because again
the tangent function is

periodic, so we need this bit.

That

And we need that and it's Mark
off this axis so we know where

we are. This is 90.

180

270 and 360. So this
must be minus 90 -

180 - 270 and minus
360.

Now 2X is Route 3, the angle
whose tangent is Route 3. We

know is 60, so we go across here
at Route 3 and we meet the curve

there and there.

And we come back this way. We
meet it there and we meet there.

So our answers are down here.

Working with this one, first we
know that that is 60, so 2X is

equal to 60 and so that that one
is 60 degrees on from that

point. Symmetry says there for
this one is also 60 degrees on

from there. In other words, it's

240. Let's work our way
backwards. This one must be 60

degrees on from minus 180, so it
must be at minus 120. This one

is 60 degrees on.

From minus 360 and so therefore
it must be minus 300.

And so if we divide throughout
by two, we have 31120 -

60 and minus 150 degrees. We
want to put degree signs on

all of these, so there are
four solutions there.

Trick equations often come up as
a result of having expressions

or other equations which are
rather more complicated than

that and depends upon

identity's. So I'm going to
have a look at a couple

of equations. These equations
both dependa pawn two identity's

that is expressions involving
trig functions that are true for

all values of X.

So the first one is sine
squared of X plus cost

squared of X is one. This is
true for all values of X.

The second one we derive from

this one. How we derive it
doesn't matter at the moment,

but what it tells us is that sex
squared X is equal to 1 + 10

squared X. So these are the two
identity's that I'm going to be

using. Sine squared X plus cost
squared X is one and sex squared

of X is 1 + 10 squared of

X OK. So how do we go
about using one of those to do

an equation like this? Cos

squared X? Plus cause
of X is equal

to sine squared of
X&X is between 180

and 0 degrees.

Well.

We've got a cost squared, A
cause and a sine squared.

If we were to use our identity
sine squared plus cost squared

is one to replace the sine
squared. Here I'd have a

quadratic in terms of Cos X, and
if I got a quadratic then I know

I can solve it either by
Factorizing or by using the

formula. So let me write down
sign squared X plus cost

squared. X is equal to 1, from
which we can see.

Sine squared X is equal to
1 minus Cos squared of X,

so I can take this and
plug it into their. So my

equation now becomes cost
squared X Plus X is equal

to 1 minus Cos squared X.

I want to get this as
a quadratic square term linear

term. Constant term equals 0, so
I begin by adding cost squared

to both sides.

So adding on a cost squared
there makes 2 Cos squared X plus

cause X equals 1. 'cause I added
cost square to get rid of that

one. Now I need to take one away
from both sides to cost squared

X Plus X minus one equals 0.

Now this is just a quadratic
equation, so the first question

I've got to ask is does it
factorize? So let's see if we

can get it to factorize.

I'll put two calls X in there
and cause X in there because

that 2 cause X times that cause
X gives Me 2 cost squared and I

put a one under one there 'cause
one times by one gives me one

and now I know to get a minus
sign. One's got to be minus and

one's got to be plus now I want
plus cause X so if I make this

one plus I'll have two cause X
times by one.

Is to cause X if I make this one
minus I'll have minus Cos X from

there. Taking those two
together, +2 cause X minus Cos X

is going to give me the plus
Kozaks in there, so that equals

0. Now, if not equal 0, I'm
multiplying 2 numbers together.

This one 2 cause X minus one and
this one cause X plus one, so

one of them or both of them have
got to be equal to 0.

So 2 calls X minus
one is 0.

All cause of X Plus One is
0, so this one tells me that

cause of X is equal to 1/2.

And this one tells Maine that
cause of X is equal to minus

one, and both of these are
possibilities. So I've got to

solve both equations to get the
total solution to the original

equation. So let's begin with
this cause of X is equal to 1/2.

And if you remember the range
of values was nought to 180

degrees, so let me sketch
cause of X between North and

180 degrees, and it looks
like that zero 9180.

We go across there at half and
come down there and there is

only one answer in the range, so
that's X is equal to 60 degrees.

But this one again let's sketch
cause of X between North and

180. There and there between
minus one and plus one and we

want cause of X equal to minus
one just at one point there and

so therefore X is equal to 180
degrees. So those are our two

answers to the full equation
that we had.

So it's now have a look at

three. 10 squared X is
equal to two sex squared X

Plus One and this time will

take X. To be between North and
180 degrees. Now, the identity

that we want is obviously the
one, the second one of the two

that we had before. In other
words, the one that tells us

that sex squared X is equal to 1
+ 10 squared X and we want to be

able to take this 1 + 10 squared
and put it into their. So we've

got 3. 10 squared
X is equal to 2

* 1 + 10 squared
X Plus one. Multiply out

this bracket. 310 squared X
is 2 + 210 squared

X plus one.

We can combine the two and the
one that will give us 3.

And we can take the 210 squared
X away from the three times

squared X there. That will give
us 10 squared X. Now we take the

square root of both sides so we
have 10X is equal to plus Route

3 or minus Route 3.

And we need to look at each of

these separately. So.

Time X equals Route

3. And Tan X
equals minus Route 3.

Access to be between North and
180, so let's have a sketch of

the graph of tan between those
values, so there is 90.

And there is 180 the angle whose
tangent is Route 3, we know.

Is there at 60 so we
know that X is equal to

60 degrees? Here we've
got minus Route 3, so

again, little sketch.

Between North and 180 range over
which were working here, we've

got minus Route 3 go across
there and down to their and

symmetry says it's got to be the
same as this one. Over here it's

got to be the same either side.
So in fact if that was 60 there

this must be 120 here, so X is
equal to 120 degrees.

So far we've been working in
degrees, but it makes little

difference if we're actually
working in radians and let's

just have a look at one or two
examples where in fact the range

of values that we've got is in
radians. So if we take Tan, X is

minus one and we take X to be
between plus or minus pie.

Another way of looking at
that would be if we were in

degrees. It will be between
plus and minus 180. Let's

sketch the graph of tangent
within that range.

Up to there.

That's π by 2.

Up to their which is π.

Minus Π
by 2.

Their minus

pie. Ton of X is
minus one, so somewhere

across here it's going to
meet the curve and we can

see that means it here.

And here giving us these
solutions at these points. Well,

we know that the angle whose
tangent is plus one is π by 4.

So this must be pie by 4 further
on, and so we have X is equal to

pie by 2 + π by 4. That will be
3/4 of Π or three π by 4, and

this one here must be.

Minus Π by 4 back there, so
minus π by 4.

Let's take one with
a multiple angle.

So we'll have a look cause
of two X is equal to Route

3 over 2.

I will take
X between North

and 2π. Now if
X is between North and 2π, and

we've got 2X.

And that means that 2X can be
between North and four π.

So again, we've got to make use
of the periodicity.

Of the graph of cosine to get a
second copy of it.

So there's the first copy
between North and 2π, and now we

want a second copy that goes
from 2π up till four π.

We can mark these off that one
will be pie by two.

Pie.

Three π by 2.

This one will be 5 Pi by
two. This one three Pi and this

one Seven π by 2.

So where are we with this cost?
2 X equals. Well, in fact we

know cost to access Route 3 over
2. We know that the angle that

gives us the cosine that is
Route 3 over 2 is π by 6. So

I'll first one is π Phi six,
root 3 over 2. Up here we go

across we meet the curve we come
down. We know that this one here

is π by 6.

Let's keep going across the
curves and see where we come to,

what we come to one here which
is π by 6 short of 2π. So

let me write it down as 2π -
Π by 6, and then again we come

to one here. Symmetry suggests
it should be pie by 6 further

on, so that's 2π + π by 6,
and then this one here.

Is symmetry would suggest his
pie by 6 short of four Pi,

so four π - π by 6.
So let's do that arithmetic 2X

is π by 6.

Now, how many sixths are there
in two? Well, the answer. Is

there a 12 of them and we're
going to take one of them away,

so that's eleven π by 6. We're
going to now add a 6th on, so

that's 13 Pi by 6.

How many 6th are there in four
or there are 24 of them? We're

going to take one away, so
that's 23. Pi over 6. Now we

want X, so we divide each of
these by 2π by 1211 Pi by

12:13, pie by 12, and 20, three
π by 12, and there are our

four solutions. Let's have a
look at one where we've got the

X divided by two rather than
multiplied by two. So the sign

of X over 2 is minus Route
3 over 2.

And let's take X to be
between pie and minus π. So

will sketch the graph of sign
between those limited, so it's

there. And their π
zero and minus pie.

Where looking for minus three
over 2. Now the one thing we do

know is that the angle who sign
is 3 over 2 is π by 3.

But we want minus Route 3 over
2, so that's down there.

We go across.

And we meet the curve these two

points. Now this curve is
symmetric with this one.

So if we know that.

Plus Route 3 over 2.

This one was Pi by three. Then
we know that this one must be

minus π by 3.

This one is π by three back, so
it's at 2π by three, so this one

must be minus 2π by three, and
so we have X over 2 is equal

to minus 2π by three and minus,
π by three, but it's X that we

want, so we multiply up X equals
minus four Pi by three and minus

2π by 3.

Let's just check on these
values. How do they fit with the

given range? Well, this 1 - 2π
by three is in that given range.

This one is outside, so we don't
want that one.

A final example here, working
with the idea again of using

those identities and will take 2
cost squared X.

Plus sign X is
equal to 1.

And we'll take X between
North and 2π.

We've got causes and signs,
so the identity that we're

going to want to help us
will be sine squared plus

cost. Squared X equals 1.

Cost squared here.

Cost squared here. Let's use
this identity to tell us that

cost squared X is equal to 1
minus sign squared X and make

the replacement up here for cost

squared. Because that as we will
see when we do it.

Leads to a quadratic in sign X,
so it's multiply this out 2 -

2 sine squared X plus sign X
is equal to 1 and I want

it as a quadratic, so I want
positive square term and then

the linear term and then the
constant term. So I need to add.

This to both sides of 0 equals 2
sine squared X. Adding it to

both sides. Now I need to take
this away minus sign X from both

sides and I need to take the two
away from both sides. So one

takeaway two is minus one.

And now does this factorize?
It's clearly a quadratic. Let's

look to see if we can make it
factorize 2 sign X and sign X.

Because multiplied together,
these two will give Me 2 sine

squared one and one because
multiplied together, these two

will give me one, but one of
them needs to be minus. To make

this a minus sign here. So I
think I'll have minus there and

plus there because two sign X
times by minus one gives me.

Minus 2 sign X one times by sign
X gives me sign X and if I

combine sign X with minus two
sign XI get minus sign X.

I have two numbers multiplied
together. This number 2 sign X

Plus One and this number sign X
minus one. They multiply

together to give me 0, so one or
both of them must be 0. Let's

write that down.

2 sign X Plus One is equal to
0 and sign X minus one is equal

to 0, so this tells me that sign
of X is equal. To take one away

from both sides and divide by
two. So sign X is minus 1/2 and

this one tells me that sign X is
equal to 1.

I'm now in a position to solve
these two separate equations.

So let me take this one first.

Now. We were working between
North and 2π, so we'll have a

sketch between North and 2π.

Of the sine curve and we want
sign X equals one. Well, there's

one and there's where it meets,
and that's pie by two, so we can

see that X is equal to pie by

two. Sign X equals minus 1/2.
Again, the range that we've been

given is between North and 2π.
So let's sketch between Norton

2π There's 2π.

Three π by 2.

Pie pie by two 0 - 1/2,
so that's coming along between

minus one and plus one that's
going to come along there.

And meet the curve there and
there. Now the one thing that we

do know is the angle who sign is

plus 1/2. Is π by 6, so we're
looking at plus 1/2. It will be

there and it would be pie by 6.

So it's π by 6 in from there,
so symmetry tells us that this

must be pie by 6 in from there,
so we've got X is equal to π

+ π by 6, and symmetry tells us
it's pie by 6 in. From there, 2π

- Π by 6.

There are six sixths in pie, so
that's Seven π by 6. There is

1216, two Pi. We're taking one
of them away, so it will be

11 Pi over 6.

So we've shown there how to
solve some trig equations.

The important thing is the
sketch the graph. Find the

initial value and then
workout where the others are

from the graphs. Remember,
the graphs are all symmetric

and they're all periodic, so
they repeat themselves every

2π or every 360 degrees.