Đáp án:
$vậy $ $x=\frac{\pi}{2},$ $x=\frac{±2\pi}{3}$
Giải thích các bước giải:
$\text{( 2cos+1)(cos2x+2sinx-2)=3-}$ $4sin^{2}x$
⇔$\text{( 2cos+1)(cos2x+2sinx-2)=}$$3-4(1-cos^{2}x)$
⇔$\text{( 2cos+1)(cos2x+2sinx-2)=}$$4cos^{2}x-1$
⇔$\text{( 2cos+1)(cos2x+2sinx-2)=}$$(2cos+1)(2cosx-1)$
⇔$\text(2cosx+1)(cos2x+2sinx-2-2cosx+1)=0$
⇔\(\left[ \begin{array}{l}2cosx+1=0\\cos2x+2(sinx-cosx)-1=0\end{array} \right.\)
$\text{xét pt: 2cosx+1=0}$
⇔$cosx=\frac{-1}{2}$
⇔$cosx=\frac{±2\pi}{3}+k2\pi$
$\text{xét pt: cos2x+2(sin-cosx)-1=0}$
⇔$-2sin^{2}x+2(sinx-cosx)=0$
⇔$sin^{2}x-sinx+cosx=0$
⇔\(\left[ \begin{array}{l}sinx=1\\cosx=0\end{array} \right.\)
⇔$sinx=sin\frac{\pi}{2}$
⇔$x=\frac{\pi}{2}+k2\pi$
