Question

Giải pt

\(6sin^23x+cos12x=4\)

Answer 1

\(\Leftrightarrow3.\left(1-cos6x\right)+2.cos^26x-1=4\)

\(\Leftrightarrow2.cos^26x-3.cos6x-2=0\)

\(\Leftrightarrow\left(cos6x-2\right)\left(2.cos6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=2\left(vn\right)\\cos6x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}6x=\dfrac{2\pi}{3}+k2\pi\\6x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\) (\(k\in Z\)) \(\Leftrightarrow\)\(\left[{}\begin{matrix}x\ne\dfrac{\pi}{9}+\dfrac{k\pi}{3}\\x\ne\dfrac{-\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\) (\(k\in Z\))

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