1: \(3\cdot cos^2x+2\cdot\sin x\cdot cosx-5\cdot\sin^2x=0\)
=>\(3\cdot\frac{1+cos2x}{2}+\sin2x-5\cdot\frac{1-cos2x}{2}=0\)
=>\(1,5+1,5\cdot cos2x+\sin2x-2,5+2,5\cdot cos2x=0\)
=>4cos2x+sin 2x-1=0
=>\(\sin2x+4\cdot cos2x=1\)
=>\(\sin2x\cdot\frac{1}{\sqrt{17}}+cos2x\cdot\frac{4}{\sqrt{17}}=\frac{1}{\sqrt{17}}\)
=>\(\sin\left(2x+\alpha\right)=cos\alpha=\sin\left(\frac{\pi}{2}-\alpha\right)\)
=>\(\left[\begin{array}{l}2x+\alpha=\frac{\pi}{2}-a+k2\pi\\ 2x+a=\pi-\frac{\pi}{2}+a+k2\pi=\frac{\pi}{2}+a+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{\pi}{2}-2a+k2\pi\\ 2x=\frac{\pi}{2}+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{4}-a+k\pi\\ x=\frac{\pi}{4}+k\pi\end{array}\right.\)
2: \(\sin^2x+2\cdot\sin x\cdot cosx-2\cdot cos^2x=\frac12\)
=>\(\frac{1-cos2x}{2}+\sin2x-2\cdot\frac{1+cos2x}{2}=\frac12\)
=>\(\frac12-\frac12\cdot cos2x+\sin2x-1-\frac12\cdot cos2x=\frac12\)
=>\(\sin2x-cos2x=1\)
=>\(\sqrt2\cdot\sin\left(2x-\frac{\pi}{4}\right)=1\)
=>\(\sin\left(2x-\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\)
=>\(\left[\begin{array}{l}2x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\ 2x-\frac{\pi}{4}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{\pi}{2}+k2\pi\\ 2x=\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{4}+k\pi\\ x=\frac{\pi}{2}+k\pi\end{array}\right.\)
