Question Type 1: Determining relationship between trigonometric functions using unit circle Exercises

Show that for all real $x$, $\sin x + \sin(-x) = 0$, and state the values of $x$ in $-2\pi < x < 2\pi$ satisfying the equation.

Determine the coordinates of the points on the unit circle corresponding to the angles $x = \tfrac{5\pi}{6}$ and $x = -\tfrac{7\pi}{6}$.

Find all values of $x$ in the interval $-2\pi < x < 2\pi$ such that $\tan x = \tan\left(\frac{\pi}{4}\right)$.

Determine all values of $x$ in $-2\pi < x < 2\pi$ for which $\cot x = 1$.

Find all values of $x$ in the interval $-2\pi < x < 2\pi$ such that $\cos x = \cos\left(\frac{\pi}{6}\right)$.

Solve $\sin x = -\tfrac{\sqrt{3}}{2}$ for $x$ in the interval $-2\pi < x < 2\pi$.

Find all values of $x$ in the interval $-2\pi < x < 2\pi$ such that $\sin x = \sin\left(\frac{\pi}{3}\right)$.

Find all values of $x$ in $-2\pi < x < 2\pi$ satisfying $\cos x = \cos\left(\tfrac{3\pi}{4}\right)$.

Determine all $x$ in $-2\pi < x < 2\pi$ satisfying $2\sin^2 x - 1 = 0$.

Solve for $x$ in $-2\pi < x < 2\pi$ the equation $\cos(2x) = \tfrac12$.

Solve $\sec x = 2$ for $x$ in the interval $-2\pi < x < 2\pi$.

Solve $\sin(2x) = \tfrac{\sqrt{2}}{2}$ for $x$ in $-2\pi < x < 2\pi$.