Question

Solve the equation for solutions in the interval 0 13 years ago

Answer 1

\[\sin 4\theta=\frac{ \sqrt{3} }{2 }=\sin \frac{ \pi }{ 3 },\sin \left( \pi-\frac{ \pi }{ 3 } \right)\]
\[=\sin \frac{ \pi }{ 3 },\sin \frac{ 2\pi }{3 }=\sin \left( 2n \pi+\frac{ \pi }{3 } \right),\sin \left( 2n \pi+\frac{ 2\pi }{3 } \right)\]
where n=0,1,2,3
\[4\theta=2n \pi+\frac{ \pi }{ 3 },2n \pi +\frac{ 2\pi }{ 3 },n=0,1,2,3\]
i think now you can solve
keep in mind
\[0\le \theta \le 2\pi hence 0 \le 4\theta \le 8 \pi \]

Answer 2

correction
\[0< \theta < 2pi \]