Question

[Châu Ngọc Minh Anh]

sin²x + cos²2x +sin²3x + cos²4x =2

Answer 1

[Hoàng Tử Hà]

\(\Leftrightarrow\sin^2x-\sin^22x+\sin^23x-\sin^24x=0\)

\(\Leftrightarrow\left(\sin x+\sin2x\right)\left(\sin2x-\sin x\right)+\left(\sin3x+\sin4x\right)\left(\sin4x-\sin3x\right)=0\)

\(\Leftrightarrow2\sin\dfrac{3}{2}x.\cos\dfrac{x}{2}.2\cos\dfrac{3}{2}x.\sin\dfrac{x}{2}+2\sin\dfrac{7}{2}x.\cos\dfrac{x}{2}.2\sin\dfrac{x}{2}\cos\dfrac{7}{2}x=0\)

\(\Leftrightarrow\sin3x.\sin x+\sin7x.\sin x=0\)

\(\Leftrightarrow\sin x\left(\sin3x+\sin7x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin x=0\\\sin3x=\sin\left(-7x\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\3x=-7x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)