Question

giải phương trình

\(8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}\)

Answer 1

ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(8sinx=\frac{\sqrt{3}sinx+cosx}{sinx.cosx}\)

\(\Leftrightarrow8sinx.cosx.sinx=\sqrt{3}sinx+cosx\)

\(\Leftrightarrow4sin2x.sinx=\sqrt{3}sinx+cosx\)

\(\Leftrightarrow2cosx+2cos2x=\sqrt{3}sinx+cosx\)

\(\Leftrightarrow cos2x=\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx\)

\(\Leftrightarrow cos2x=sin\left(x-\frac{\pi}{6}\right)=cos\left(\frac{2\pi}{3}-x\right)\)

\(\Leftrightarrow.......\)