# 5. Solving Trigonometric Equations

by M. Bourne

Trigonometric equations can be solved using the algebraic methods and trigonometric identities and values discussed in earlier sections. You may wish to go back and have a look at Trigonometric Functions of Any Angle, where we see the background to the following solutions.

A painless way to solve these is using a graph. Where the graph cuts the *x*-axis, that's where you'll find your solutions (the *x*-values that "work"). Graphs also help you to understand why sometimes there is one answer, and sometimes many answers. I use Scientific Notebook or similar math software to graph the functions for me.

You can use this Online Graphing Calculator to solve the following equations (or check your solutions) .

### Example 1

Solve the equation 2 cos *θ* − 1 = 0 for 0 ≤ *θ* < 2*π*.

Answer

Rearranging the above equation, we get:

`cos theta=1/2`

We know the following :

Since

`cos (pi/3) = 1/2`

and `cos θ` is positive in the first and fourth quadrants, we have:

`theta=pi/3`

or

`theta=2pi-pi/3=(5pi)/3`

So `theta=pi/3` or `theta=(5pi)/3`

### Example 2

### Revision Tip

Getting lost in this section? See the background at Trigonometric Functions of Any Angle

Graphically solve the equation

2 cos^{2}x− sinx− 1 = 0

such that 0 ≤ *θ* < 2*π*.

Answer

Using graphing software, we draw the curve of *y* = 2 cos^{2}*x* − sin *x* − 1 in the region 0 ≤ *θ* < 2π. Wherever the curve cuts the *x*-axis will be the solution for our equation.

We see from the graph that the solutions are approximately:

*x* = 0.5

*x* = 2.6 *
x* = 4.7

For more accurate solutions, we would just zoom in on the graph.

In this next graph, I have zoomed in to the second root (the one near *x* = 2.6). We can see that this root is *x* = 2.618 correct to 3 decimal places.

We could continue to zoom in as close as we like to get the accuracy that we require.

## Solving Equations Involving Multiples of *θ*

### Example 3

Solve the
equation sin 2*θ* = 0.8 for 0 ≤ *θ* < 2π.

Answer

If the problem involved *θ* only, we would expect 2 solutions; one
in the first quadrant and one in the second quadrant.

But here our problem involves `2θ`, so we have to **double**
the domain (*θ* values) to account for all
possible solutions.

We proceed as follows:

We solve

`sin 2θ = 0.8` for 0 ≤ 2

θ< 4π.

The reference angle is

`α = arcsin 0.8 = 0.9273`

So the values for 2*θ* will be in quadrants I, II, V, VI.

2

θ= 0.9273, orπ− 0.9273, or 2π+ 0.9273, or 3π− 0.9273

That is

`2θ = 0.9273, 2.2143, 7.2105, 8.4975`

But we need values for *θ*, not 2*θ*,
so we divide throughout by 2:

`θ = 0.4637, 1.1072, 3.6053, 4.2488`

Are our answers correct? As usual, we will check by graphing the original expression:

We can see from the graph that our 4 values are reasonable, since these are the only 4 values that satisfy `sin 2θ = 0.8`.

**Note: **We can
always check our solutions with calculator, but it is easy to "miss out" on some of the required values if we only
use calculator.

### Example 4

Solve the equation

`cos^2theta=1/16`

for 0 ≤ *θ* < 2π.

Answer

Solving for cos *θ* gives us:

`cos theta=+-1/4`

If `cos alpha=1/4`, then the reference angle is α = 1.3181.

So for `cos theta=1/4`, we have *θ* in the first and 4th quadrants. So

`θ = 1.3181 or 4.9651`

For `cos theta=-1/4`, we have *θ* in the 2nd and 3rd quadrants. So

`θ = 1.8235 or 4.4597`

So ` θ= 1.3181, 1.8235,` ` 4.4597` ` or 4.9651` radians.

We can see from the graph of `y=cos^2theta-1/16` that our answer is correct:

### Example 5

Solve the equation

6 sin

^{2}θ− sinθ− 1 = 0

for 0 ≤ *θ* < 2*π*.

Answer

Factoring the LHS:

`6 sin^2θ − sin θ − 1 = 0`

`(2 sin θ − 1)(3 sin θ + 1) = 0`

So either

`2 sin θ − 1 = 0`

`sin θ = 1/2`

θwill be in 1st and 2nd quadrants.`θ = 0.52360, 2.6180` (that is, `pi/6` or `(5pi)/6`)

OR

`3 sin θ + 1 = 0`

`sin θ = − 1/3`

θwill be in 3rd and 4th quadrants.`θ = 3. 4814, 5. 9433`

Checking our solution on a graph:

So `θ = 0.52360, 2.6180, 3. 4814, 5. 9433`

### Example 6

Solve the equation

`cos{:x/2:}=1+cos\ x`

for 0 ≤ *θ* < 2*π*.

Answer

Earlier we learned that `cos(x/2)=+-sqrt((1+cos x)/2)`, so we have:

`+-sqrt((1+cos x)/2)= 1 + cos x`

Squaring both sides gives:

`(1+cos x)/2=(1+cos x)^2`

`(1+cos x)/2=1+2\ cos x+cos^2x`

`1+cos x=2+4\ cos x+2\ cos^2x`

So we have:

`2\ cos^2 \ x + 3\ cos x + 1 = 0``(2\ co\s x + 1)(cos x + 1) = 0`

Solving, we get

`cos x = − 0.5` or `cos x = − 1`

Now `cos x=-1/2` gives `x=(2pi)/3,(4pi)/3`.

However, on checking in the **original **equation,
we note that

`"LHS"=cos((4pi)/3xx1/2)` `=cos((2pi)/3)` `=-1/2`

but

`"RHS"=1+cos (4pi)/3=+1/2`

So the only solution for this part is `x=(2pi)/3.`

Also, `cos x=-1` gives `x = pi`.

So the solutions for the equation are `x=(2pi)/3or pi.`

A check of the graph of `y=cos x/2-1-cos x` confirms these results:

(2π/3 ≈ 2.0944 and π ≈ 3.14).

### Example 7

Solve the equation

tan 2

θ− cot 2θ= 0

for 0 ≤ *θ* < 2π.

Answer

`cot 2theta= 1/(tan 2 theta)`, so we have:

`tan 2 theta-1/(tan 2 theta)=0`

`tan^2 2θ = 1`

`tan 2θ = ± 1`

Since `0 ≤ θ < 2π`, we need to consider values of `2theta` such that `0 ≤ 2θ < 4π`. Hence, solving the above equation, we have:

`2 theta=pi/4,(3pi)/4,(5pi)/4(7pi)/4,` `(9pi)/4,(11pi)/4,` `(13pi)/4,(15pi)/4`

Dividing throughout by 2 gives us the full set of solutions in the required domain, `0 <= theta <2pi`:

`theta=pi/8,(3pi)/8,` `(5pi)/8,(7pi)/8,` `(9pi)/8,(11pi)/8,` `(13pi)/8,(15pi)/8`

## Exercises

Note 1:"Analytically" means use the methods and formulas from previous sections. It means don't just use a graph to solve it.

Note 2:However, I always use a graph to check my analytical work. I can see immediately if some error has occurred. I encourage you to do the same!

**1.** Solve the trigonometric equation analytically

4 tan

x− sec^{2}x= 0 (for 0 ≤x< 2π)

Answer

`4\ tan x− sec^2x= 0`

Writing this in terms of `sin x` and `cos x` only:

`4(sin x)/(cos x)-1/(cos^2x)=0`

Multiplying throughout by `cos x`:

`4\ sin x\ cos x=1`

Dividing both sides by 2:

`2\ sin x\ cos x=1/2`

Recognizing the LHS is `sin 2x`, from before:

`sin 2x,=0.5`

In 0 ≤ *x* < 2π, we need to
find values of 2*x* such that
0 ≤ 2*x* < 4π. (Twice the original domain.)

So the values for `2x` are:

`2x=pi/6,(5pi)/6,(13pi)/6,(17pi)/6`

Dividing throughout by 2 gives our required values for `x`:

`x=pi/12,(5pi)/12,(13pi)/12,(17pi)/12`

or, in decimal form:

`x = 0.2618, 1.309, 3.403, 4.451`

**2.** Solve the trigonometric equation
analytically for 0 ≤ *x* < 2*π*:

sin 2

xcosx− cos 2xsinx= 0

Answer

We recognise the left hand side to be in the form:

`sin(a − b) =` ` sin a cos b − cos a sin b,`

where `a = 2x`* *and `b =x`.

So

`sin 2x\ cos x − cos 2x\ sin x`

`= sin(2x − x)`

`= sin x`

Now, we know the solutions of `sin x = 0` to be:

`x = 0, π`.

[Why?]

### Need Graph Paper?

**3.** Solve the given trigonometric equation
analytically and by graphical method (for 0 ≤ *x* < 2*π*):

sin 4

x− cos 2x= 0

Answer

sin 4*x* − cos 2*x* = 0

2sin 2*x *cos 2*x* − cos 2*x*
= 0

Factoring gives:

cos 2*x* (2 sin 2*x *− 1) = 0

EITHER

`cos 2x= 0`

`2x=pi/2,(3pi)/2,(5pi)/2,(7pi)/2`

`x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4`

OR

`sin 2x= 1/2`

`2x=pi/6,(5pi)/6,(13pi)/6,(17pi)/6`

`x=pi/12,(5pi)/12,(13pi)/12,(17pi)/12`

Or, in decimal form:

`x= 0.26, 0.79, ` `1.31, 2.36, ` `3.40, 3.93, ` `4.45, 5.50.`

The graph of `y = sin 4x− cos 2x` is as follows. We can see from where the graph cuts the *x*-axis that our answers are reasonable.