Question

Giải phương trình:

\(1+\cos4x-2\sin^{2\left(x\right)}=0\)

Answer 1

Đề là thế này: \(1+cos4x-2sin^2x=0\)

Hay thế này: \(1+cos4x-2sin2x=0\)

Answer 2

\(1+cos4x-2sin2x=0\)

\(\Leftrightarrow1+1-2sin^22x-2sin2x=0\)

\(\Leftrightarrow sin^22x+sin2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{5}-1}{2}=sin\alpha\\sin2x=\frac{-\sqrt{5}-1}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\alpha+k2\pi\\2x=\pi-\alpha+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\alpha}{2}+k\pi\\x=\frac{\pi}{2}-\frac{\alpha}{2}+k\pi\end{matrix}\right.\)

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