Harder trigonometric equations

In a nutshell

By adjusting the given range, it is possible to solve equations of the form $\sin(n\theta) = p$ , $\cos(n\theta) = p$ and $\tan(n\theta) = p$ . Similar adjustments can also be used to solve equations of the form $\sin(\theta + \alpha) = p$ , $\cos(\theta + \alpha)=p$ and $\tan(\theta + \alpha) = p$ .

Simple trigonometric equations with $n\theta$

You have already learned that equations of the form $\sin(\theta) = p$ and $\cos(\theta) = p$ have solutions in a given range when $-1 \leq p \leq 1$ . It follows that equations of the form $\sin(n\theta) = p$ and $\cos(n\theta) =p$ also have solutions when $-1\leq p \leq1$ .

Now, consider finding the solutions to an equation of the form $\sin(n\theta) = p$ , $\cos(n\theta) = p$ or $\tan(n\theta) = p$ over some interval $D \leq \theta \leq D'$ . Notice that if $D \leq \theta \leq D'$ then it follows that $nD \leq n\theta \leq nD'$ . Therefore, equations with $n\theta$ can be solved by multipying up the range for $\theta$ given in the question. The final values for $\theta$ can then be found by dividing all the solutions in the range $nD \leq n\theta \leq nD'$ by $n$ to give $D \leq \theta \leq D'$ .

Example 1

Find all solutions to the equation $\sin(2\theta) = \dfrac12$ in the interval $0\degree \leq \theta \leq 360\degree$ .

First multiply up the range. As the trig function has $2\theta$ , multiply the range by $2$ to give:

$0\degree \leq 2\theta \leq 720\degree$ 

This means the solutions to $\sin(2\theta) = \dfrac12$ for $2\theta$ must lie in the range $0\degree \leq 2\theta \leq 720\degree$ .

Find the principal value:

$\begin{aligned}\sin(2\theta) &= \dfrac12 \\2\theta&= \sin^{-1} \Big( \dfrac {1} {2} \Big ) \\2 \theta &= 30 \degree\end{aligned}$ 

Using a $\sin$ graph or otherwise, find that the full set of solutions in the interval $0\degree \le 2\theta \le 720\degree$ :

$2\theta = 30\degree, \underbrace{150\degree}_{180 - 30}, \underbrace{390\degree}_{30+360}$ and $\underbrace{510\degree}_{150+360}$ .

Dividing through by $n = 2$ , the full set of solutions to $\sin(2\theta) = \dfrac12$ in the range $0\degree \leq \theta \leq 360\degree$ is given by:

$\dfrac{30\degree}{2}= 15\degree$ , $\dfrac{150\degree}{2} = 75\degree$ , $\dfrac{390\degree}{2} = 195\degree$  and $\dfrac{510\degree}{2} = 255\degree$ .

Therefore, the full set of solutions to the equation $\sin(2\theta) = \dfrac12$ in the interval $0\degree \le \theta \le 360\degree$ is given by $\underline{\theta=15\degree, 75\degree, 195\degree, 255\degree}$ .

Simple trigonometric equations with $\theta + \alpha$

Similar to the above, for $D \leq \theta \le D'$ then $D + \alpha \le \theta + \alpha \le D'+\alpha$ . Therfore, when given equations of the form $\sin(\theta + \alpha)= p$ , $\cos(\theta + \alpha)=p$ or $\tan(\theta + \alpha)=p$ , find solutions in the range $D + \alpha \le \theta + \alpha \le D'+\alpha$ before subtracting $\alpha$ . This will results in the range for $\theta$ being $D \leq \theta \le D'$ .

Example 2

Find all solutions to the equation $\cos(\theta + 45\degree) = \dfrac57$ in the interval $0\degree \leq \theta \le 360\degree$ . Give your answers to $1$ decimal place.

First, adjust the range of $\theta$ by adding $45\degree$ to range $0\degree \leq \theta \le 360\degree$ to give the interval $45\degree \leq \theta+45\degree \leq 405\degree$ .

Find the principal value:

$\begin{aligned}\cos(\theta + 45\degree) &= \dfrac57 \\\theta + 45 \degree&= \cos^{-1}\Big( \dfrac 5 7 \Big) \\\theta + 45 \degree &= 44.415\dots \degree\end{aligned}$

This lies outside the desired range! Using a $\cos$ graph or otherwise, deduce that the full set of solutions lying in the range $45^\degree \leq \theta+45\degree \leq 405\degree$ is given by:

$\theta + 45 \degree = 44.415 \degree, \underbrace{315.585 \degree}_{360-44.415}, \underbrace{404.415 \degree}_{44.415+360}$ 

Subtracting $45\degree$ from the solutions gives:

$\theta = 270.6 \degree, 359.4 \degree$ 

Therefore, the full set of solutions to the equation $\cos(\theta + 45\degree) = \dfrac57$ , to $1$ decimal place, in the interval $0\degree \le \theta \le 360\degree$ , is given by $\underline{\theta = 270.6\degree}$ and $\underline{\theta = 359.4\degree}$ .