Q.Find the general solution of the equation $(\sqrt{3} - 1)\cos\theta + (\sqrt{3} + 1)\sin\theta = 2$.

A trigonometric equation like $\sin x = 0$ or $\cos x = 1/2$ has infinitely many solutions, because every trig function repeats itself every $2\pi$ (or $\pi$, for $\tan$). Instead of listing solutions one by one, the goal is a single formula -- the general solution -- that generates every one of them as an integer $n$ ranges over all of $\mathbb{Z}$.

The three base formulas, each proved directly from where the function is zero or repeats:

• $\sin x = 0 \iff x = n\pi$
• $\cos x = 0 \iff x = (2n+1)\dfrac{\pi}{2}$
• $\tan x = 0 \iff x = n\pi$

The three general formulas, for a known value $\sin x = \sin y$ (and likewise for $\cos$, $\tan$), where $y$ is any one known solution (often found from a standard angle):

• $\sin x = \sin y \iff x = n\pi + (-1)^n y$
• $\cos x = \cos y \iff x = 2n\pi \pm y$
• $\tan x = \tan y \iff x = n\pi + y$

Worked example. Solve $\sin x = \dfrac{1}{2}$.

Since $\sin(\pi/6) = 1/2$, take $y = \pi/6$. By the sine formula, the general solution is $x = n\pi + (-1)^n \dfrac{\pi}{6}$, $n \in \mathbb{Z}$. Checking $n=0$ gives $x=\pi/6$ (correct, $\sin(\pi/6)=1/2$); $n=1$ gives $x = \pi - \pi/6 = 5\pi/6$ (also correct, since $\sin(5\pi/6) = \sin(\pi/6) = 1/2$) -- the alternating $(-1)^n$ is exactly what reflects the solution into the second quadrant on odd $n$. …