h: \(3\cdot cos^2x+\sin2x-\sin^2x=2\)
=>\(4\cdot cos^2x-\sin^2x-cos^2x+\sin2x=2\)
=>\(4\cdot cos^2x+\sin2x-3=0\)
=>\(4\cdot\frac{1+cos2x}{2}+\sin2x-3=0\)
=>\(2+2\cdot cos2x+\sin2x-3=0\)
=>\(2\cdot cos2x+\sin2x-1=0\)
=>\(2\cdot cos2x+\sin2x=1\)
=>\(cos2x\cdot\frac{2}{\sqrt5}+\sin2x\cdot\frac{1}{\sqrt5}=\frac{1}{\sqrt5}\)
=>\(\sin\left(\alpha+2x\right)=cos\alpha=\sin\left(\frac{\pi}{2}-\alpha\right)\)
=>\(\left[\begin{array}{l}2x+\alpha=\frac{\pi}{2}-\alpha+k2\pi\\ 2x+\alpha=\pi-\frac{\pi}{2}+\alpha+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=-2a+\frac{\pi}{2}+k2\pi\\ 2x=\frac{\pi}{2}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\alpha+\frac{\pi}{4}+k\pi\\ x=\frac{\pi}{4}+k\pi\end{array}\right.\)
