IB Mathematics Analysis and Approaches 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 $0\le x<2\pi$: $$\sin (2x)=\sqrt{3}\cos (x)$$

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