1. **Problem statement:** Solve the following trigonometric equations for $x$. 2. **Recall the general solutions:** - For $\sin x = a$, solutions are $x = \arcsin(a) + 2k\pi$ or $x = \pi - \arcsin(a) + 2k\pi$. - For $\cos x = a$, solutions are $x = \arccos(a) + 2k\pi$ or $x = -\arccos(a) + 2k\pi$. - For $\tan x = a$, solutions are $x = \arctan(a) + k\pi$. Here, $k$ is any integer. 3. **Solve each equation:** (i) $\sin x = \frac{1}{2}$ - $x = \frac{\pi}{6} + 2k\pi$ or $x = \frac{5\pi}{6} + 2k\pi$ (ii) $\sin x = -\frac{\sqrt{2}}{2}$ - $x = -\frac{\pi}{4} + 2k\pi$ or $x = \frac{5\pi}{4} + 2k\pi$ (iii) $\sin x = 0$ - $x = k\pi$ (iv) $\sin x = \frac{1}{3}$ - $x = \arcsin\left(\frac{1}{3}\right) + 2k\pi$ or $x = \pi - \arcsin\left(\frac{1}{3}\right) + 2k\pi$ (v) $\sin x = -\frac{1}{4}$ - $x = -\arcsin\left(\frac{1}{4}\right) + 2k\pi$ or $x = \pi + \arcsin\left(\frac{1}{4}\right) + 2k\pi$ (vi) $\cos x = \frac{1}{2}$ - $x = \frac{\pi}{3} + 2k\pi$ or $x = -\frac{\pi}{3} + 2k\pi$ (vii) $\cos x = -\frac{\sqrt{3}}{2}$ - $x = \frac{5\pi}{6} + 2k\pi$ or $x = \frac{7\pi}{6} + 2k\pi$ (viii) $\cos x = -1$ - $x = \pi + 2k\pi$ (ix) $\cos x = \frac{1}{3}$ - $x = \arccos\left(\frac{1}{3}\right) + 2k\pi$ or $x = -\arccos\left(\frac{1}{3}\right) + 2k\pi$ (x) $\cos x = -\frac{1}{4}$ - $x = \arccos\left(-\frac{1}{4}\right) + 2k\pi$ or $x = -\arccos\left(-\frac{1}{4}\right) + 2k\pi$ (xi) $\tan x = \sqrt{3}$ - $x = \frac{\pi}{3} + k\pi$ (xii) $\tan x = 1$ - $x = \frac{\pi}{4} + k\pi$ (xiii) $\tan x = 2$ - $x = \arctan(2) + k\pi$ 4. **Explanation:** - The solutions use the inverse trigonometric functions to find principal values. - Because sine and cosine are periodic with period $2\pi$, and tangent with period $\pi$, we add multiples of these periods. - For sine and cosine, there are two solutions per period due to symmetry. - For tangent, solutions repeat every $\pi$. 5. **Triangle context:** - The triangle with sides 1 (adjacent), $\sqrt{3}$ (opposite), and 2 (hypotenuse) corresponds to angles of $30^\circ$ ($\pi/6$) and $60^\circ$ ($\pi/3$), which appear in the solutions above. Final answers are expressed with $k \in \mathbb{Z}$ for all integer multiples of the periods.