Question

1. 2cos2x + sinx = sin3x

2. cos²x + 2(sin3x-1)sin²(π/4 - x/2) = 0

Answer 1

1.

\(\Leftrightarrow2cos2x+sinx-sin3x=0\)

\(\Leftrightarrow2cos2x-2cos2x.sinx=0\)

\(\Leftrightarrow2cos2x\left(1-sinx\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

Answer 2

2.

\(cos^2x+\left(sin3x-1\right)\left(1-cos\left(\dfrac{\pi}{2}-x\right)\right)=0\)

\(\Leftrightarrow1-sin^2x+\left(sin3x-1\right)\left(1-sinx\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx\right)+\left(sin3x-1\right)\left(1-sinx\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx+sin3x-1\right)=0\)

\(\Leftrightarrow2\left(1-sinx\right)sin2x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sin2x=0\\cosx=0\end{matrix}\right.\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\dfrac{k\pi}{2}\)