Question

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

Answer 1

đk \(X\ne\dfrac{k\pi}{2}\left(k\in Z\right)\)

\(8sinx.cos^2x=\sqrt{3}cosx+sinx\)

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

\(\Leftrightarrow4.\dfrac{1}{2}\left(sin3x+sinx\right)=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow2sin3x+2sinx=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow2sin3x=\sqrt{3}cosx-sinx\)

\(\Leftrightarrow sin3x=\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx\)

\(\Leftrightarrow sin3x=sin\left(\dfrac{\pi}{3}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{3}-x+k2\pi\\3x=x+\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.\left(k\in Z\right)\)