IB Mathematics Applications & Interpretation Question Type 1: Determining Relationship Between Trigonometric Functions Using Unit Circle Exercises

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

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 $x=\frac{5\pi }{6}$ and $x=-\frac{7\pi }{6}$.

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

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

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

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

Solve for $x$ in $-2\pi <x<2\pi$ the equation $\cos (2x)=\frac{1}{2}$.

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

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

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

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