Question

Giải phương trình

Answer 1

ĐKXĐ: \(sinx\ne0\)

Chia 2 vế cho \(sin^2x\) ta được:

\(\dfrac{3}{sin^2x}-4=4cotx.\left(\dfrac{cosx}{sinx}\right)^2-\dfrac{3cotx}{sin^2x}\)

\(\Leftrightarrow3\left(1+cot^2x\right)-4=4cotx.cot^2x-3cotx\left(1+cot^2x\right)\)

\(\Leftrightarrow3cot^2x-1=4cot^3x-3cotx-3cot^3x\)

\(\Leftrightarrow cot^3x-3cot^2x-3cotx+1=0\)

\(\Leftrightarrow\left(cotx+1\right)\left(cot^2x-4cotx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2-\sqrt{3}\\cotx=2+\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{5\pi}{12}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)