Question Type 6: Solving more difficult equations that require rearranging to solving for x Bootcamps

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

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

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

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

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

Solve for $x$ in the equation

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

giving the general solution.

Solve for $x$ in $0 \le x < 2\pi$: $\cos x\,(\tfrac{1}{2} + \sin 2x) = \dfrac{\sin x}{\tan x}$.

Solve for $x$ in the equation

$\sin x + \sin2x = 0,$

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

Solve for $x$ in the equation

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

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

Solve for $x$ in $0 \le x < 2\pi$: $\sin\!\big(x + \tfrac{\pi}{3}\big) = \cos(2x)$.

Solve for $x$ in the equation

$\cos x\left(\frac12 + \sin2x\right) = \frac{\sin x}{\tan x},$

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

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