Question

8sin2x - 12sinxcos^2x + 3cosx = 4

Answer 1

\(8sin2x-12sinx.cos^2x+3cosx=4\)

\(\Leftrightarrow8sin2x-6.\left(2sinx.cosx\right).cosx+3cosx-4=0\)

\(\Leftrightarrow8sin2x-6sin2x.cosx+3cosx-4=0\)

\(\Leftrightarrow2sin2x\left(4-3cosx\right)-\left(4-3cosx\right)=0\)

\(\Leftrightarrow\left(4-3cosx\right)\left(2sin2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{4}{3}>1\left(vn\right)\\sin2x=\dfrac{1}{2}\end{matrix}\right.\)

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