IB Mathematics Analysis and Approaches Question Type 6: Solving More Difficult Equations That Require Rearranging to Solving for x Exercises

Solve $\sin x+\sin 2x=0$ for $0\le x<2\pi$.

Solve for $x$ in the equation

$$\cos x\left(\frac{1}{2}+\sin 2x\right)=\frac{\sin x}{\tan x},$$

where $x$ is in the interval $[0,2\pi )$.

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

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

Solve for $x$ in the equation

$$\sin x+\sin 2x=0,$$

where $x$ is in $[0,2\pi )$.

Solve for $x$ in $0\le x<2\pi$: $\tan x+\sin 2x=0$.

Solve for $x$ in the equation

$$3{\cos }^{2}x-4\sin x=1$$

for $x$ in $[0,2\pi )$.

Solve for $x$ in the equation

$$\tan (2x)=\sqrt{3},$$

giving the general solution.

Solve for $x$ in $0\le x<2\pi$: $2{\cos }^{2}x-\sin x=1$.

Solve for $x$ in the interval $0\le x<2\pi$.

Solve for $x$ in $0\le x<2\pi$: $\sin \left(x+\frac{\pi }{3}\right)=\cos (2x)$.

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