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

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

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

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

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

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

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 $0\le x<2\pi$: $\sin(4x)=2\sin(2x).$

Solve for $x$ in $0\le x<2\pi$: $\cos(2x)+\cos(4x)=\tfrac12.$

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

Solve for $x$ in $0\le x<2\pi$: $3\sin(x)-4\sin^3(x)=\tfrac12.$

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