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 $$\underline{\theta=15\degree, 75\degree, 195\degree, 255\degree}$$.

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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 $$\underline{\theta = 359.4\degree}$$.