Solve the equation for solutions in the interval [0, 2 π).
-$2 \sqrt { 3 } \sin 4 x = 3$

Question

Quizplus · Accepted Answer

First, divide both sides by $2\sqrt{3}$ to isolate $\sin 4x$, giving $\sin 4x = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$. The solutions to $\sin \theta = \frac{\sqrt{3}}{2}$ are $\theta = \frac{\pi}{3}$ and $\theta = \frac{2\pi}{3}$ within the interval $[0, 2\pi)$. Since we have $4x$ instead of $\theta$, we solve $4x = \frac{\pi}{3} + 2\pi k$ and $4x = \frac{2\pi}{3} + 2\pi k$ for $x$, where $k$ is an integer. This gives multiple solutions within the interval $[0, 2\pi)$, specifically $x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{9\pi}{12} (= \frac{3\pi}{4}), \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{21\pi}{12} (= \frac{7\pi}{4})$, but considering the symmetry and periodicity of the sine function, we adjust to find all unique solutions in the given interval, leading to the set in option C.

Solve the Equation for Solutions in the Interval [0, 2 23sin⁡4x=32 \sqrt { 3 } \sin 4 x = 323​sin4x=3

Solve the Equation for Solutions in the Interval [0, 2 $2 \sqrt { 3 } \sin 4 x = 3$