Question

\(\sin^22x-\sin^2x=\frac{1}{2}\)

\(2\sin^2x+4\sin x=3\cos^2x\)

\(\sin^3x+3\sin^2x+2\sin x=0\)

Answer 1

a.

\(1-cos^22x-\left(\frac{1-cos2x}{2}\right)=\frac{1}{2}\)

\(\Leftrightarrow2cos^22x-cos2x=0\)

\(\Leftrightarrow cos2x\left(2cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=\frac{1}{2}\\\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)

Answer 2

b.

\(\Leftrightarrow2sin^2x+4sinx=3\left(1-sin^2x\right)\)

\(\Leftrightarrow5sin^2x+4sinx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{-2-\sqrt{19}}{5}\left(l\right)\\sinx=\frac{-2+\sqrt{19}}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{-2+\sqrt{19}}{5}\right)+k2\pi\\x=\pi-arcsin\left(\frac{-2+\sqrt{19}}{5}\right)+k2\pi\end{matrix}\right.\)

c.

\(\Leftrightarrow sinx\left(sin^2x+3sinx+2\right)=0\)

\(\Leftrightarrow sinx\left(sinx+1\right)\left(sinx+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)