Question Type 3: Using identities to solve difficult trigonometric equations Exercises

Solve for $x$ in $0<x<\pi$: $\cot^2(x)-\csc(x)\cot(x)=0.$

Solve for $x$ in the interval $0 \le x < 2\pi$: $\sin(x) + \sin(2x) - \sin(3x) = 0.$

$1+\tan x=\sec x$

Solve for $x$ in $0 \le x < 2\pi$: $\cos(2x) + \cos(4x) = \frac{1}{2}$

Solve for $x$ in $0 \le x < 2\pi$: $\sin(3x) = \cos(x)$

Solve for $x$ in the interval $0\le x<2\pi$: $\sin(2x)=\sqrt3\cos(x)$

Solve for $x$ in $0\le x<2\pi$: $\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=0.$

Solve for $x$ in $0 < x < \frac{\pi}{2}$: $\tan^2(x)-2\tan(x)\sec(x)+1=0.$

Solve for $x$ in $0\le x<2\pi$: $2\sin^2(x)-3\sin(x)\cos(x)-\cos^2(x)=0.$

Solve for $x$ in the interval $0 \le x < 2\pi$: $3\sin(x)-4\sin^3(x)=\frac{1}{2}$

Solve for $x$ in $0\le x<2\pi$: $\sin(4x)=2\sin(2x)$

Solve for $x$ in the interval $0<x<\pi$: $\frac{\sin(3x)}{\sec^2(x)}=\sin(x)\cos^2(x)+\sin^2(x)\cos^2(x)+\cos^4(x)-\frac{1}{\sec^2(x)}.$