lm trên symbolab.com

\(\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\cos^2x\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\left(2-\sin^2x\right)\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=4\sin^2x-1\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=\left(2\sin x-1\right)\left(2\sin x+1\right)\)

\(\Leftrightarrow2\sin2x+1=2\sin x+1\)

\(\Leftrightarrow\sin2x=\sin x\)

\(\Leftrightarrow\sin2x-\sin x=0\)

\(\Leftrightarrow2\cos\frac{3}{2}-\cos\frac{x}{2}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\cos\frac{3}{2}=0\\\cos\frac{x}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{cases}\left(k\inℤ\right)}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{3}+\frac{2\pi}{3}k\\x=\pi+4k\pi\end{cases}\left(k\inℤ\right)}\)

MIK MỚI LỚP 7. THÔNG CẢM

ko trl thì thôi

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

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

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

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

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

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

\(\eta=Pi;3.14159\)

những câu hỏi không liên quan đến THCS thì bạn vào h để có thể được giải đáp tốt hơn

"h" nhé mình đánh thiếu