Trigonometric Equations and Principal Solutions

A solution of a trigonometric equation is any value of the angle that makes the equation true when substituted in. For example, $\theta=\dfrac{\pi}{6}$ satisfies $\sin\theta=\dfrac12$ because $\sin\dfrac{\pi}{6}=\dfrac12$; so does $\theta=\dfrac{5\pi}{6}$, because $\sin\dfrac{5\pi}{6}=\dfrac12$ as well. In fact the same equation is also satisfied by $\theta=\dfrac{5\pi}{6}+2\pi=\dfrac{17\pi}{6}$, by $\theta=\dfrac{\pi}{6}-2\pi=-\dfrac{11\pi}{6}$, and so on forever, purely because sine has period $2\pi$. Likewise $\theta=\dfrac{\pi}{4}$ satisfies $\cos\theta=\dfrac{1}{\sqrt2}$, and $\theta=\dfrac{7\pi}{4}$ satisfies $\cos\theta=\dfrac{1}{\sqrt2}$ too (since $\cos\dfrac{7\pi}{4}=\cos\left(2\pi-\dfrac{\pi}{4}\right)=\cos\dfrac{\pi}{4}$).

Because of this periodicity a trigonometric equation can have infinitely many solutions, so mathematicians agree to first describe only the solutions that lie in one standard window, $[0,2\pi)$ — one full period, starting from $0$ and not including $2\pi$ itself.

Definition (Principal Solution). A solution $\alpha$ of a trigonometric equation is called a principal solution if $0\le\alpha<2\pi$.

For example, both $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$ lie in $[0,2\pi)$, so both are principal solutions of $\sin\theta=\dfrac12$. But $\dfrac{13\pi}{6}$, although it does satisfy $\sin\theta=\dfrac12$ (because $\dfrac{13\pi}{6}=2\pi+\dfrac{\pi}{6}$), is not a principal solution, since $\dfrac{13\pi}{6}\notin[0,2\pi)$.

A function's value can also fix how many principal solutions an equation has: $\theta=0$ is the only principal solution of $\sin\theta=0$ in $[0,2\pi)$ (note $2\pi$ itself is excluded by the strict upper bound), while $\cos\theta=-1$ has exactly one principal solution, $\theta=\pi$, because cosine only touches $-1$ once per period.

Worked Example 1 — Find the principal solutions of $\sin\theta=\dfrac12$. Since $\sin\dfrac{\pi}{4}$... (correction — using the standard angle) $\sin\dfrac{\pi}{6}=\dfrac12$ and $0\le\dfrac{\pi}{6}<2\pi$, so $\dfrac{\pi}{6}$ is one principal solution. Using the allied-angle identity $\sin\theta=\sin(\pi-\theta)$, we also get $\sin\left(\pi-\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{6}=\dfrac12$, i.e. $\sin\dfrac{5\pi}{6}=\dfrac12$, and $\dfrac{5\pi}{6}$ also lies in $[0,2\pi)$. So the two principal solutions are $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$.

Worked Example 2 — Find the principal solutions of $\cos\theta=\dfrac12$. $\cos\dfrac{\pi}{3}=\dfrac12$ and $\dfrac{\pi}{3}\in[0,2\pi)$, so $\dfrac{\pi}{3}$ is one principal solution. Using $\cos\theta=\cos(2\pi-\theta)$: $\cos\left(2\pi-\dfrac{\pi}{3}\right)=\cos\dfrac{\pi}{3}=\dfrac12$, i.e. $\cos\dfrac{5\pi}{3}=\dfrac12$, and $\dfrac{5\pi}{3}\in[0,2\pi)$. So the principal solutions are $\dfrac{\pi}{3}$ and $\dfrac{5\pi}{3}$. …