The General Solution

Since a trigonometric equation has infinitely many solutions repeating every period, we want one formula — parametrised by an integer $n$ — that generates every solution. This is the general solution. For instance, all solutions of $\sin\theta=\dfrac12$ are $\ldots,-\dfrac{7\pi}{6},\dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{13\pi}{6},\ldots$, and every one of these is generated by the single expression $n\pi+(-1)^n\dfrac{\pi}{6}$, $n\in\mathbb{Z}$: this expression is called the general solution of $\sin\theta=\dfrac12$.

Theorem 3.1 (general solution of $\sin\theta=\sin\alpha$). The general solution of $\sin\theta=\sin\alpha$ is $\theta=n\pi+(-1)^n\alpha$, $n\in\mathbb{Z}$.

Proof. Since $\sin\theta=\sin\alpha$, $\alpha$ is one solution. Since $\sin(\pi-\alpha)=\sin\alpha$, $\pi-\alpha$ is a second solution. Using the $2\pi$-periodicity of sine on each of these: $\sin\theta=\sin\alpha=\sin(2\pi+\alpha)=\sin(4\pi+\alpha)=\cdots$, and separately $\sin\theta=\sin(\pi-\alpha)=\sin(3\pi-\alpha)=\sin(5\pi-\alpha)=\cdots$. So every solution is either of the form (even multiple of $\pi$) $+\,\alpha$, or of the form (odd multiple of $\pi$) $-\,\alpha$. Listing them in order: $\theta=\ldots,\alpha,\ \pi-\alpha,\ 2\pi+\alpha,\ 3\pi-\alpha,\ 4\pi+\alpha,\ 5\pi-\alpha,\ldots$ — the sign on $\alpha$ alternates and the coefficient of $\pi$ increases by 1 each time, which is exactly captured by $\theta=n\pi+(-1)^n\alpha$ for $n\in\mathbb{Z}$ (even $n$ gives $+\alpha$, odd $n$ gives $-\alpha$). $\blacksquare$

Theorem 3.2 (general solution of $\cos\theta=\cos\alpha$). The general solution of $\cos\theta=\cos\alpha$ is $\theta=2n\pi\pm\alpha$, $n\in\mathbb{Z}$.

Proof. $\alpha$ is one solution. Since $\cos(-\alpha)=\cos\alpha$, $-\alpha$ is a second solution. By periodicity, $\cos\theta=\cos\alpha=\cos(2\pi+\alpha)=\cos(4\pi+\alpha)=\cdots$ and $\cos\theta=\cos(-\alpha)=\cos(2\pi-\alpha)=\cos(4\pi-\alpha)=\cdots$. So $\theta=\alpha,\,2\pi+\alpha,\,4\pi+\alpha,\ldots$ or $\theta=-\alpha,\,2\pi-\alpha,\,4\pi-\alpha,\ldots$, which together are exactly $\theta=2n\pi\pm\alpha$, $n\in\mathbb{Z}$. $\blacksquare$

Theorem 3.3 (general solution of $\tan\theta=\tan\alpha$). The general solution of $\tan\theta=\tan\alpha$ is $\theta=n\pi+\alpha$, $n\in\mathbb{Z}$.

Proof. $\tan\theta=\tan\alpha$ iff $\dfrac{\sin\theta}{\cos\theta}=\dfrac{\sin\alpha}{\cos\alpha}$ iff $\sin\theta\cos\alpha=\cos\theta\sin\alpha$ iff $\sin\theta\cos\alpha-\cos\theta\sin\alpha=0$ iff $\sin(\theta-\alpha)=0=\sin0$. By the general solution of $\sin\theta=\sin\alpha$ (Theorem 3.1, with $0$ in place of $\alpha$), $\theta-\alpha=n\pi+(-1)^n\cdot 0=n\pi$, i.e. $\theta=n\pi+\alpha$, $n\in\mathbb{Z}$. $\blacksquare$

Remark (zero-crossing cases). For $\theta\in\mathbb{R}$: (i) $\sin\theta=0$ iff $\theta=n\pi$, $n\in\mathbb{Z}$; (ii) $\cos\theta=0$ iff $\theta=(2n+1)\dfrac{\pi}{2}$, $n\in\mathbb{Z}$; (iii) $\tan\theta=0$ iff $\theta=n\pi$, $n\in\mathbb{Z}$.

Theorem 3.4 (general solution of $\sin^2\theta=\sin^2\alpha$). The general solution is $\theta=n\pi+\alpha$, $n\in\mathbb{Z}$.

Proof (first method). $\sin^2\theta=\sin^2\alpha\implies\sin\theta=\pm\sin\alpha\implies\sin\theta=\sin\alpha$ or $\sin\theta=\sin(-\alpha)$. By Theorem 3.1, $\theta=n\pi+(-1)^n\alpha$ or $\theta=n\pi+(-1)^n(-\alpha)$; in either case the set of solutions collapses to $\theta=n\pi+\alpha$, $n\in\mathbb{Z}$.

Proof (second method, via double angle). $\sin^2\theta=\sin^2\alpha\implies\dfrac{1-\cos2\theta}{2}=\dfrac{1-\cos2\alpha}{2}\implies\cos2\theta=\cos2\alpha$. By Theorem 3.2, $2\theta=2n\pi+2\alpha$, so $\theta=n\pi+\alpha$, $n\in\mathbb{Z}$.

Theorem 3.5 (general solution of $\cos^2\theta=\cos^2\alpha$). The general solution is $\theta=n\pi+\alpha$, $n\in\mathbb{Z}$.

Proof. $\cos^2\theta=\cos^2\alpha\implies\dfrac{1+\cos2\theta}{2}=\dfrac{1+\cos2\alpha}{2}\implies\cos2\theta=\cos2\alpha\implies2\theta=2n\pi+2\alpha\implies\theta=n\pi+\alpha$, $n\in\mathbb{Z}$.

Theorem 3.6 (general solution of $\tan^2\theta=\tan^2\alpha$). The general solution is $\theta=n\pi+\alpha$, $n\in\mathbb{Z}$.

Proof. $\tan^2\theta=\tan^2\alpha\implies\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$ (componendo-dividendo) $\implies\cos2\theta=\cos2\alpha\implies2\theta=2n\pi+2\alpha\implies\theta=n\pi+\alpha$, $n\in\mathbb{Z}$.

Solved Examples.

Ex.(1) Find the general solution of (i) $\sin\theta=\dfrac{\sqrt3}{2}$ (ii) $\cos\theta=\dfrac12$ (iii) $\tan\theta=\sqrt3$.

1. $\sin\theta=\sin\dfrac{\pi}{3}\implies\theta=n\pi+(-1)^n\dfrac{\pi}{3}$, $n\in\mathbb{Z}$ (Theorem 3.1).
2. $\cos\theta=\cos\dfrac{\pi}{4}\implies\theta=2n\pi+\dfrac{\pi}{4}$, $n\in\mathbb{Z}$ (Theorem 3.2, single-sign form since the equation is satisfied exactly at $+\pi/4$ here).
3. $\tan\theta=\tan\dfrac{\pi}{3}\implies\theta=n\pi+\dfrac{\pi}{3}$, $n\in\mathbb{Z}$ (Theorem 3.3). Ex.(2) Find the general solution of (i) $\sin\theta=-\dfrac{\sqrt3}{2}$ (ii) $\cos\theta=-\dfrac12$ (iii) $\cot\theta=-\sqrt3$.

(i) Since $\sin\dfrac{4\pi}{3}=-\dfrac{\sqrt3}{2}$ (as $\sin(\pi+A)=-\sin A$), $\theta=n\pi+(-1)^n\dfrac{4\pi}{3}$, $n\in\mathbb{Z}$.

(ii) Since $\cos\dfrac{2\pi}{3}=-\dfrac12$ (as $\cos(\pi-A)=-\cos A$), $\theta=2n\pi\pm\dfrac{2\pi}{3}$, $n\in\mathbb{Z}$.

(iii) $\cot\theta=-\sqrt3\implies\tan\theta=-\dfrac{1}{\sqrt3}$; since $\tan\dfrac{5\pi}{6}=-\dfrac{1}{\sqrt3}$ (as $\tan(\pi-A)=-\tan A$), $\theta=n\pi+\dfrac{5\pi}{6}$, $n\in\mathbb{Z}$.

Ex.(3) Find the general solution of (i) $\text{cosec}\,\theta=2$ (ii) $\sec\theta+\sqrt2=0$.

(i) $\text{cosec}\,\theta=2\implies\sin\theta=\dfrac12=\sin\dfrac{\pi}{6}\implies\theta=n\pi+(-1)^n\dfrac{\pi}{6}$, $n\in\mathbb{Z}$.

(ii) $\sec\theta=-\sqrt2\implies\cos\theta=-\dfrac{1}{\sqrt2}=\cos\dfrac{3\pi}{4}$ (as $\cos(\pi-A)=-\cos A$) $\implies\theta=2n\pi+\dfrac{3\pi}{4}$, $n\in\mathbb{Z}$.

Ex.(4) Find the general solution of (i) $\cos2\theta=-\dfrac12$ (ii) $\tan3\theta=-1$ (iii) $\sin4\theta=\dfrac{\sqrt3}{2}$.

(i) $\cos2\theta=\cos\dfrac{3\pi}{4}$ (since $\cos\dfrac{3\pi}4=-\dfrac1{\sqrt2}$... corrected: using $\cos\dfrac{3\pi}{4}=-\dfrac{1}{\sqrt2}$ does not equal $-\frac12$; instead use $\cos\dfrac{2\pi}{3}=-\dfrac12$) so $\cos2\theta=\cos\dfrac{2\pi}{3}\implies2\theta=2n\pi\pm\dfrac{2\pi}{3}\implies\theta=n\pi\pm\dfrac{\pi}{3}$, $n\in\mathbb{Z}$.

(ii) $\tan3\theta=\tan\dfrac{3\pi}{4}$ (as $\tan\dfrac{3\pi}4=-1$) $\implies3\theta=n\pi+\dfrac{3\pi}{4}\implies\theta=\dfrac{n\pi}{3}+\dfrac{\pi}{4}$, $n\in\mathbb{Z}$.

(iii) $\sin4\theta=\sin\dfrac{\pi}{3}\implies4\theta=n\pi+(-1)^n\dfrac{\pi}{3}\implies\theta=\dfrac{n\pi}{4}+(-1)^n\dfrac{\pi}{12}$, $n\in\mathbb{Z}$.

Ex.(5) Find the general solution of (i) $4\cos^2\theta=1$ (ii) $4\sin^2\theta=3$ (iii) $\tan^2\theta=1$.

(i) $\cos^2\theta=\dfrac14=\cos^2\dfrac{\pi}{3}\implies\theta=n\pi\pm\dfrac{\pi}{3}$, $n\in\mathbb{Z}$ (Theorem 3.5).

(ii) $\sin^2\theta=\dfrac34=\sin^2\dfrac{\pi}{3}\implies\theta=n\pi\pm\dfrac{\pi}{3}$, $n\in\mathbb{Z}$ (Theorem 3.4).

(iii) $\tan^2\theta=1=\tan^2\dfrac{\pi}{4}\implies\theta=n\pi+\dfrac{\pi}{4}$, $n\in\mathbb{Z}$ (Theorem 3.6).

Ex.(6) Find the general solution of $\cos3\theta=\cos2\theta$. $\cos3\theta-\cos2\theta=0\implies-2\sin\dfrac{5\theta}{2}\sin\dfrac{\theta}{2}=0\implies\sin\dfrac{5\theta}{2}=0$ or $\sin\dfrac{\theta}{2}=0\implies\dfrac{5\theta}{2}=n\pi$ or $\dfrac{\theta}{2}=n\pi$, $n\in\mathbb{Z}\implies\theta=\dfrac{2n\pi}{5}$ or $\theta=2n\pi$. Since $2n\pi$ is a subset of $\dfrac{2n\pi}{5}$ (take multiples of 5), the combined required general solution is $\theta=\dfrac{2n\pi}{5}$, $n\in\mathbb{Z}$.

Ex.(7) Find the general solution of $\cos5\theta=\sin3\theta$. $\cos5\theta=\cos\left(\dfrac{\pi}{2}-3\theta\right)\implies5\theta=2n\pi\pm\left(\dfrac{\pi}{2}-3\theta\right)$. Taking the minus sign: $5\theta=2n\pi-\dfrac{\pi}{2}+3\theta\implies2\theta=2n\pi-\dfrac{\pi}{2}\implies\theta=n\pi-\dfrac{\pi}{4}$. Taking the plus sign: $5\theta=2n\pi+\dfrac{\pi}{2}-3\theta\implies8\theta=2n\pi+\dfrac{\pi}{2}\implies\theta=\dfrac{n\pi}{4}+\dfrac{\pi}{16}$. So $\theta=n\pi-\dfrac{\pi}{4}$ or $\theta=\dfrac{n\pi}{4}+\dfrac{\pi}{16}$, $n\in\mathbb{Z}$. …