Question

[Nguyễn Đức Hiếu]

2sin2x-cos2x=7sinx+2cosx-4

Answer 1

[✎﹏🅷ạ🅽🅷︵❣🅿🅷ú🅲︵❣Đé🅾︵❣🅲ó︵❣Đâ🆄︵❣✔]

\(2\sin2x-\cos2x=7\sin x+2\cos x-4\)

\(\Rightarrow4\sin x\cos x-\left(1-2\sin^2x\right)-7\sin x-2\cos x+4=0\)

\(\Rightarrow2\cos x\left(2\sin x-1\right)+\left(2\sin^2-7\sin x+3\right)=0\)

\(\Rightarrow2\cos x\left(2\sin x-1\right)+\left(2\sin x-1\right)\left(\sin x-3\right)=0\)

\(\Rightarrow\left(2\sin x-1\right)\left(2\cos x+\sin x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2\sin x-1=0\\2\cos x+\sin x=3,\left(vn\right)\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{\eta}{6}+k2\eta\\x=\frac{5\eta}{6}+k2\eta\end{cases}}}\)

\(\eta=Pi;3.14159\)