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

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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 $\frac{2\pi}{3}$ within the first cycle of the sine function. Since the argument of the sine function is $4x$, we divide the standard solutions by 4 to find the solutions within the interval $[0, 2\pi)$. This process yields eight solutions, accounting for the periodicity of the sine function and the fact that it repeats every $\pi$ radians when multiplied by 4, fitting within the specified interval.

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

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