# A calculus of the absurd

#### 7.4 General solutions to trigonometric equations

When is this equation true?

\begin{equation*} \sin (x) = \frac {\sqrt {3}}{2} \hspace {5pt} 0 < x < 2\pi \end{equation*}

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

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

We can call the value returned by any inverse trig function the “principle 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 $$\sin (\theta ) = a$$

\begin{align} \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} \end{align}

And that for $$\cos (\theta )$$

\begin{align} \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} \end{align}

This is also equivalent to

\begin{align} \theta = \pm \arccos (a) \pm 2\pi \cdot k \hspace {12pt} k \in \mathbb {N} \end{align}