Solve the equation on the interval [0, 2π).
-$\sin 4 x = \frac { \sqrt { 3 } } { 2 }$
A) $\frac { \pi } { 12 } , \frac { \pi } { 6 } , \frac { 2 \pi } { 3 } , \frac { 7 \pi } { 12 } , \frac { 7 \pi } { 6 } , \frac { 13 \pi } { 12 } , \frac { 5 \pi } { 3 } , \frac { 19 \pi } { 12 }$
B) $\frac { \pi } { 4 } , \frac { 5 \pi } { 4 }$
C) $0 , \frac { \pi } { 4 } , \pi$
D) 0

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Quizplus · Accepted Answer

The equation $\sin 4x = \frac{\sqrt{3}}{2}$ has solutions where the sine function equals $\frac{\sqrt{3}}{2}$, which occurs at angles $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ in the first revolution (0 to $2\pi$). Since the argument of the sine function is $4x$, we divide these angles by 4 to find the solutions in terms of $x$, and consider all possible angles within the interval [0, $2\pi$) that would satisfy the equation when multiplied by 4. This results in eight specific solutions, correctly listed in option A.

Solve the Equation on the Interval [0, 2π) sin⁡4x=32\sin 4 x = \frac { \sqrt { 3 } } { 2 }sin4x=23​​

Solve the Equation on the Interval [0, 2π) $\sin 4 x = \frac { \sqrt { 3 } } { 2 }$