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

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

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 $0 \leq x < 2\pi$: $\cos x \cos 2x = \sin x \sin 2x$.