Câu hỏi của Julian Edward

Question

Giải các pt

a) $\frac{2cosx-1}{2cosx+1}=3$

b) $cosx\left(2cos2x-1\right)=3cosx$

c) $sin2x.cos2x=0$

Nguyễn Việt Lâm · Accepted Answer

a/ ĐKXĐ: $cosx\ne-\frac{1}{2}$

$\Leftrightarrow2cosx-1=6cosx+3$

$\Leftrightarrow4cosx=-4\Rightarrow cosx=-1$

$\Rightarrow x=\pi+k2\pi$

b/

$\Leftrightarrow cosx\left(2cos2x-1\right)-3cosx=0$

$\Leftrightarrow cosx\left(2cos2x-4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}cosx=0\cos2x=2\left(vn\right)\end{matrix}\right.$

$\Rightarrow x=\frac{\pi}{2}+k\pi$

c/

$\Leftrightarrow2sin2x.cos2x=0$

$\Leftrightarrow sin4x=0$

$\Rightarrow4x=k\pi\Rightarrow x=\frac{k\pi}{4}$Câu trả lời: a/ ĐKXĐ: $cosx\ne-\frac{1}{2}$