Because of the periodic nature of trigonometric functions, a trigonometric equation will usually have many solutions and these will occur at regular intervals. If we can find one or more basic solutions, the others can be specified by adding multiples of some fixed amount to the known solutions.

For example, the equation $\sin\theta=\frac{1}{\sqrt{3}}$sinθ=1√3 has the solution $\theta_1=\arcsin\frac{1}{\sqrt{3}}$θ1=arcsin1√3 in the first quadrant - between $0$0 and $\frac{\pi}{2}$π2. It must also have the second quadrant solution $\theta_2=\pi-\arcsin\frac{1}{\sqrt{3}}$θ2=π−arcsin1√3.

We know that the sine function is periodic with period $2\pi$2π. Thus, any multiple of $2\pi$2π added to the first or second quadrant solution will also be a solution. This means further solutions can be expressed as

$\theta_1+2n\pi$θ1+2nπ and

$\theta_2+2n\pi$θ2+2nπ which is the same as $\pi-\theta_1+2n\pi=(2n+1)\pi-\theta_1$π−θ1+2nπ=(2n+1)π−θ1, where $n$n is any positive or negative integer.

Guidance about finding initial solutions to trigonometric equations can be found at linked chapters A and B.

Examples

SINE

Find the general solution of $\sin\theta=\frac{1}{2}$sinθ=12 in radians.

You may use $n$n to represent an integer.

We are being asked to find all the values for $\theta$θ, that make $\sin\theta=\frac{1}{2}$sinθ=12 true.

Let's just remind ourselves of the graphical solution.

Now - let's see how to do this analytically (which just means by hand!)

Firstly we should recognise that $\sin\frac{\pi}{6}=\frac{1}{2}$sinπ6=12.

So we should use $\frac{\pi}{6}$π6 as a reference angle to find all angles in the interval $\left[0,2\pi\right)$[0,2π) whose sine is equal to $\frac{1}{2}$12.

In the interval $\left[0,2\pi\right)$[0,2π), $\theta$θ can be both $\frac{\pi}{6}$π6 and $\frac{5\pi}{6}$5π6.

Since $\sin\theta$sinθ has a period of $2\pi$2π, then the general solution will be

$\theta=\alpha+2n\pi$θ=α+2nπ or $\theta=\beta+2n\pi$θ=β+2nπ

where $\alpha$α and $\beta$β are all the solutions in the interval $\left[0,2\pi\right)$[0,2π)

So $\alpha$α is $\frac{\pi}{6}$π6 and $\beta$β is $\frac{5\pi}{6}$5π6

Which means our general solutions are

$\theta_1=\frac{\pi}{6}+2n\pi$θ1=π6+2nπ and $\theta_2=\frac{5\pi}{6}+2n\pi$θ2=5π6+2nπ

Let's just check the graph again, how these general solutions map.

Tangent

Suppose that in the course of an investigation, we need to solve the following trigonometric equation:

$3\sin2x=4\cos^2x$3sin2x=4cos2x

We might begin by writing $\sin2x$sin2x as $2\sin x\cos x$2sinxcosx. Then, the equation is equivalent to $6\sin x\cos x=4\cos^2x$6sinxcosx=4cos2x.

On moving all the terms to the left and factoring, we have, $\cos x\left(6\sin x-4\cos x\right)=0$cosx(6sinx−4cosx)=0.

This is true when $\cos x=0$cosx=0 and when $\left(6\sin x-4\cos x\right)=0$(6sinx−4cosx)=0.

From the factor $\cos x=0$cosx=0, we have $x=\frac{\pi}{2}$x=π2 or $x=-\frac{\pi}{2}$x=−π2, and when we add $2n\pi$2nπ to each of these solutions because of the periodicity of the cosine function, we get the general solutions $x=2n\pi+\frac{\pi}{2}$x=2nπ+π2 and $x=2n\pi-\frac{\pi}{2}$x=2nπ−π2 In this case, the difference between successive solutions is just $\pi$π. That is, $2n\pi-\frac{\pi}{2}+\pi=2n\pi+\frac{\pi}{2}$2nπ−π2+π=2nπ+π2 and also, $2n\pi+\frac{\pi}{2}+\pi=2(n+1)\pi-\frac{\pi}{2}$2nπ+π2+π=2(n+1)π−π2. So, we can write the solutions more simply as

$x=\frac{\pi}{2}+n\pi$x=π2+nπ.

From the other factor, $6\sin x-4\cos x=0$6sinx−4cosx=0, we can write a simpler equation $6\frac{\sin x}{\cos x}=4$6sinxcosx=4. Then, simplifying further, $\tan x=\frac{2}{3}$tanx=23.

Therefore, $x=\arctan\frac{2}{3}$x=arctan23 in the first quadrant. The tangent function has period $\pi$π. So, the set of solutions arising from this one is given by

$x=\arctan\frac{2}{3}+n\pi$x=arctan23+nπ, for integers $n$n.

We could, for practical purposes, give an approximate evaluation of the $\arctan$arctan expression. Thus, $x\approx0.5880+3.1416\ n$x≈0.5880+3.1416 n. Again for practicality, we might wish to convert this to degrees. Thus, $x\approx0.588\times\frac{360^\circ}{2\pi}+180^\circ\ n\ \approx33.6899^\circ+180^\circ\ n$x≈0.588×360°2π+180° n ≈33.6899°+180° n.

The locations of the solutions can be seen in the following graph.

cosine

Find all the solutions of the following equation: $\sin^2x+\left(\frac{\sqrt{3}-1}{2}\right)\cos x=1-\frac{\sqrt{3}}{4}$sin2x+(√3−12)cosx=1−√34.

This looks difficult but we take it in small steps.

We use the Pythagorean identity to obtain $1-\cos^2x+\left(\frac{\sqrt{3}-1}{2}\right)\cos x=1-\frac{\sqrt{3}}{4}$1−cos2x+(√3−12)cosx=1−√34.

Then, after collecting like terms and reversing the signs, $\cos^2x+\left(\frac{1-\sqrt{3}}{2}\right)\cos x=\frac{\sqrt{3}}{4}$cos2x+(1−√32)cosx=√34. We see that the equation is a quadratic equation in $\cos x$cosx. We could resort immediately to the quadratic formula, or perhaps rewrite it in factored form as

$\left(\cos x+\frac{1}{2}\right)\left(\cos x-\frac{\sqrt{3}}{2}\right)=0$(cosx+12)(cosx−√32)=0

From here, we see that $\cos x=-\frac{1}{2}$cosx=−12 and $\cos x=\frac{\sqrt{3}}{2}$cosx=√32 are solutions to the quadratic equation.

Hence, in the second and third quadrants, $x=\pi-\frac{\pi}{3}$x=π−π3 and $x=\pi+\frac{\pi}{3}$x=π+π3, while in the first and fourth quadrants, $x=\frac{\pi}{6}$x=π6 and $x=2\pi-\frac{\pi}{6}$x=2π−π6.

Putting this together, we can write the general solution as

$x=\frac{2\pi}{3}+2n\pi$x=2π3+2nπ or $x=\frac{4\pi}{3}+2n\pi$x=4π3+2nπ or $x=\frac{\pi}{6}+2n\pi$x=π6+2nπ or $x=\frac{11\pi}{6}+2n\pi$x=11π6+2nπ.

The right- and left-hand sides of the equation have been graphed. The solutions are where the graphs intersect.

Worked Examples

Question 1

Question 2

Question 3

Find the general solution of $\tan2x=\cot x$tan2x=cotx in radians.