Question

(2sinx +1)(3cos4x +2sinx -4)+4cos²x=3

Answer 1

\(6sinx.cos4x+4sin^2x-8sinx+3cos4x+2sinx-4+4cos^2x=3\)

\(\Leftrightarrow6sinx.cos4x-6sinx+3cos4x-3=0\)

\(\Leftrightarrow cos4x\left(2sinx+1\right)-\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left(cos4x-1\right)\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)