7.4 General solutions to trigonometric equations

When is this equation true?

sin(x)=320<x<2π\sin(x)=\frac{\sqrt{3}}{2}\hskip 5.0pt0<x<2\pi

Using the spangles (in the previous section), we know that it’s true when x=π3x=\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)\sin(x), it’s clear that there’s another solution to this in the interval (π2,π)\left(\frac{\pi}{2},\pi\right). Because of sin\sin’s symmetry, we know that this solution will be at ππ3\pi-\frac{\pi}{3}.

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

sin(θ)=aθ=arcsin(a)\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)\sin(x) and cos(x)\cos(x) repeat every 2π2\pi (or 360360^{\circ}) so for any value of θ\theta which solves sin(θ)=a\sin(\theta)=a or cos(θ)=a\cos(\theta)=a, then that value plus or minus any multiple of 2π2\pi (or 360360^{\circ}) will also solve the equation.

The other thing to bear in mind is that sin(θ)\sin(\theta) has an axis of symmetry in the lines x=π2x=\frac{\pi}{2} and x=3π2x=\frac{3\pi}{2}. This means that if θ\theta is the principal value solving sin(θ)=a\sin(\theta)=a, then sin(πθ)=a\sin(\pi-\theta)=a is also a solution.

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

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

θ={arcsin(a)±2πkπarcsin(a)±2πkk\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(θ)\cos(\theta)

θ={arccos(a)±2πk2πarccos(a)±2πkk\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

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