Question

Giài pt : 8cos2x = \(\dfrac{\sqrt{3}}{sinx}\)+ \(\dfrac{1}{cosx}\)

Answer 1

ĐK: \(x\ne k\pi;x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

\(8cos2x=\dfrac{\sqrt{3}}{sinx}+\dfrac{1}{cosx}\Rightarrow8cos2x.sinx.cosx=\sqrt{3}cosx+sinx\)

<=>4cos2x.sin2x=\(\sqrt{3}\)cosx+sinx

<=>2cos4x=\(\sqrt{3}\)cosx+sinx

<=>cos4x=\(\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)

<=>cos4x=cos\(\dfrac{\pi}{6}\).cosx+sin\(\dfrac{\pi}{6}\).sinx

<=>cos4x=cos(\(\dfrac{\pi}{6}\)-x)

<=>\(\left[{}\begin{matrix}4x=\dfrac{\pi}{6}-x+k2\pi\\4x=\pi-\left(\dfrac{\pi}{6}-x\right)+k2\pi\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=\dfrac{\pi}{30}+k\dfrac{2\pi}{5}\\x=\dfrac{5\pi}{18}+k\dfrac{2\pi}{3}\end{matrix}\right.\)