Question

\(\frac{2\sin^2x-\cos x-1}{t\text{anx}-\sqrt{3}}=0\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\tanx\ne\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne\frac{\pi}{3}+k\pi\end{matrix}\right.\)

Pt tương đương:

\(2sin^2x-cosx-1=0\)

\(\Leftrightarrow2\left(1-cos^2x\right)-cosx-1=0\)

\(\Leftrightarrow-2cos^2x-cosx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\frac{\pi}{3}+k2\pi\left(l\right)\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

Answer 2

bạn ơi, sao phải loại tanx-căn 3 vậy