Question

(2sinx+1)² + (2sinx +1) (sinx -\(\frac{3}{2}\))= 0

Answer 1

\(\Leftrightarrow\left(2sinx+1\right)\left(2sinx+1+sinx-\frac{3}{2}\right)=0\)

\(\Leftrightarrow\left(2sinx+1\right)\left(3sinx-\frac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=\frac{1}{6}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=arcsin\left(\frac{1}{6}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{6}\right)+k2\pi\end{matrix}\right.\)