General Solutions of Trigonometric Equations

Because every trigonometric function is periodic (Section 5), an equation

such as $\sin\theta=\sin\alpha$ is never satisfied by a single value of $\theta$ alone -- once

one solution $\alpha$ is known, so is every coterminal angle, and (for sine) so is every angle

whose terminal side is the mirror image of $\alpha$'s across the y-axis. The general solution is the formula that captures every one of these values at once, using an arbitrary

integer $n$.

Theorem 1. $\sin\theta=\sin\alpha$ if and only if $\theta=n\pi+(-1)^n\alpha$ for some

integer $n$.

Proof. $\sin\theta-\sin\alpha=0$. By the Section 8 sum-to-product formula,

so either $\cos\left(\dfrac{\theta+\alpha}{2}\right)=0$ or $\sin\left(\dfrac{\theta-\alpha}{2}\right)=0$.

Case 1: $\sin\left(\dfrac{\theta-\alpha}{2}\right)=0 \Rightarrow \dfrac{\theta-\alpha}{2}=m\pi \Rightarrow \theta = 2m\pi+\alpha$, for any integer $m$ -- these are exactly the values

$\theta=n\pi+(-1)^n\alpha$ with $n=2m$ even, since $(-1)^{2m}=1$.

Case 2: $\cos\left(\dfrac{\theta+\alpha}{2}\right)=0 \Rightarrow \dfrac{\theta+\alpha}{2}= (2m+1)\dfrac{\pi}{2} \Rightarrow \theta = (2m+1)\pi-\alpha$, for any integer $m$ -- these are

exactly the values $\theta=n\pi+(-1)^n\alpha$ with $n=2m+1$ odd, since $(-1)^{2m+1}=-1$ gives

$\theta=(2m+1)\pi-\alpha$.

Together, every solution in both cases is captured by the single formula

$\theta=n\pi+(-1)^n\alpha$, $n\in\mathbb{Z}$. $\blacksquare$

Theorem 2. $\cos\theta=\cos\alpha$ if and only if $\theta=2n\pi\pm\alpha$ for some integer

$n$.

Proof. $\cos\theta-\cos\alpha=0$. By the Section 8 formula,

so either $\sin\left(\dfrac{\theta+\alpha}{2}\right)=0$ or $\sin\left(\dfrac{\theta-\alpha}{2}\right)=0$.

Case 1: $\sin\left(\dfrac{\theta-\alpha}{2}\right)=0 \Rightarrow \theta=2n\pi+\alpha$.

Case 2: $\sin\left(\dfrac{\theta+\alpha}{2}\right)=0 \Rightarrow \theta=2n\pi-\alpha$.

Together, $\theta=2n\pi\pm\alpha$, $n\in\mathbb{Z}$. $\blacksquare$

Theorem 3. $\tan\theta=\tan\alpha$ if and only if $\theta=n\pi+\alpha$ for some integer $n$.

Proof. $\tan\theta=\tan\alpha \Leftrightarrow \dfrac{\sin\theta}{\cos\theta}= \dfrac{\sin\alpha}{\cos\alpha} \Leftrightarrow \sin\theta\cos\alpha-\cos\theta\sin\alpha=0 \Leftrightarrow \sin(\theta-\alpha)=0$ (Section 6, difference formula), which holds exactly

when $\theta-\alpha=n\pi$, i.e. $\theta=n\pi+\alpha$, $n\in\mathbb{Z}$. $\blacksquare$

Method for a general trigonometric equation. A compound equation is first reduced, using

the identities of Sections 4, 6-9, to one of these three basic forms $\sin\theta=\sin\alpha$

(or $\cos\theta=\cos\alpha$ or $\tan\theta=\tan\alpha$) -- or a product of several such

equalities set to zero -- and only then is the matching theorem applied. The reduction step …