# 7.4 General solutions to trigonometric equations

When is this equation true?

$\sin(x)=\frac{\sqrt{3}}{2}\hskip 5.0pt0

Using the spangles (in the previous section), we know that it’s true when $x=\frac{\pi}{3}$. In the interval in question, however, this isn’t the only point where it’s true! If you look at the graph of $\sin(x)$, it’s clear that there’s another solution to this in the interval $\left(\frac{\pi}{2},\pi\right)$. Because of $\sin$’s symmetry, we know that this solution will be at $\pi-\frac{\pi}{3}$.

We can generalise this to any point (and for different trig functions). For any point $a$ which is in the range of $\sin(x)$, we know

$\sin(\theta)=a\implies\theta=\arcsin(a)$ (7.25)

We can call the value returned by any inverse trig function the \sayprinciple value. It is usually the angle closest to $0$. However, this value is not necessarily the only possible value. For one, we know that $\sin(x)$ and $\cos(x)$ repeat every $2\pi$ (or $360^{\circ}$) so for any value of $\theta$ which solves $\sin(\theta)=a$ or $\cos(\theta)=a$, then that value plus or minus any multiple of $2\pi$ (or $360^{\circ}$) will also solve the equation.

The other thing to bear in mind is that $\sin(\theta)$ has an axis of symmetry in the lines $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$. This means that if $\theta$ is the principal value solving $\sin(\theta)=a$, then $\sin(\pi-\theta)=a$ is also a solution.

For $\cos(\theta)$, something very similar is the case, except that the axis of symmetry is in the line $x=\pi$, and thus if $\theta$ is the principal value solving $\cos(\theta)=a$, then $\cos(2\pi-\theta)=a$ is also a solution.

Overall, we can write that for $\sin(\theta)=a$

 \displaystyle\theta=\begin{cases}\begin{aligned} &\arcsin(a)\pm 2\pi\cdot k\\ &\pi-\arcsin(a)\pm 2\pi\cdot k\end{aligned}&k\in\mathbb{N}\end{cases} (7.26)

And that for $\cos(\theta)$

 \displaystyle\theta=\begin{cases}\begin{aligned} &\arccos(a)\pm 2\pi\cdot k\\ &2\pi-\arccos(a)\pm 2\pi\cdot k\end{aligned}&k\in\mathbb{N}\end{cases} (7.27)

This is also equivalent to

 $\displaystyle\theta=\pm\arccos(a)\pm 2\pi\cdot k\hskip 12.0ptk\in\mathbb{N}$ (7.28)