Cho phương trình: sinx.cos4x−sin^22x=4sin^2(π/4−x/2)−7/2

Hướng dẫn giải:

Ta có: (*) \( \Leftrightarrow \sin x.\cos 4x-\frac{1}{2}(1-\cos 4x)=2\left[ 1-\cos \left( \frac{\pi }{2}-x \right) \right]-\frac{7}{2} \)

 \( \Leftrightarrow \sin x\cos 4x-\frac{1}{2}+\frac{1}{2}\cos 4x=-\frac{3}{2}-2\sin x\Leftrightarrow \sin x\cos 4x+\frac{1}{2}\cos 4x+1+2\sin x=0 \)

 \( \Leftrightarrow \cos 4x\left( \sin x+\frac{1}{2} \right)+2\left( \sin x+\frac{1}{2} \right)=0\Leftrightarrow (\cos 4x+2)\left( \sin x+\frac{1}{2} \right)=0 \)

\(\Leftrightarrow \left[ \begin{align}  & \cos 4x=-2\text{ }(\ell ) \\  & \sin x=-\frac{1}{2}=\sin \left( -\frac{\pi }{6} \right) \\ \end{align} \right.\)\(\Leftrightarrow \left[ \begin{align}  & x=-\frac{\pi }{6}+k2\pi  \\  & x=\frac{7\pi }{6}+h2\pi  \\ \end{align} \right.\).

Ta có:  \( \left| x-1 \right|<3\Leftrightarrow -3<x-1<3\Leftrightarrow -2<x<4 \).

+ Với \(x=-\frac{\pi }{6}+k2\pi \), ta có:  \( -2<-\frac{\pi }{6}+k2\pi <4\Leftrightarrow -2+\frac{\pi }{6}<k2\pi <4+\frac{\pi }{6} \)

 \( \Leftrightarrow \frac{1}{12}-\frac{1}{\pi }<k<\frac{2}{\pi }+\frac{1}{12} \).

Do  \( k\in \mathbb{Z} \) nên k = 0. Vậy  \( x=-\frac{\pi }{6} \).

+ Với \(x=\frac{7\pi }{6}+h2\pi \), ta có: \(-2<\frac{7\pi }{6}+h2\pi <4\Leftrightarrow -2-\frac{7\pi }{6}<h2\pi <4-\frac{7\pi }{6}\)

\(\Leftrightarrow -\frac{1}{\pi }-\frac{7}{12}<h<\frac{2}{\pi }-\frac{7}{12}\xrightarrow{h\in \mathbb{Z}}h=0\).  Vậy  \( x=\frac{7\pi }{6} \).

Kết luận:  \( x\in \left\{ -\frac{\pi }{6};\frac{7\pi }{6} \right\} \).