Question

Giải phương trình sau: \(8\sin x\sin2x+6\sin\left(x+\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}-2x\right)=5+7\cos x\)

Answer 1

\(\Leftrightarrow8sinx.sin2x+3\left(sinx+cosx\right)\left(sin2x+cos2x\right)=5+7cosx\)

\(\Leftrightarrow8sinx.sin2x+3\left(cos2x.cosx+sin2x.sinx\right)+3\left(sinx.cos2x+cosx.sin2x\right)=5+7cosx\)

\(\Leftrightarrow8sinx.sin2x+3cosx+3sin3x=5+7cosx\)

\(\Leftrightarrow4cosx-4cos3x+3cosx+3sin3x=5+7cosx\)

\(\Leftrightarrow3sin3x-4cos3x=5\)

\(\Leftrightarrow\frac{3}{5}sin3x-\frac{4}{5}cos3x=1\)

Đặt \(cosa=\frac{3}{6}\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow sin3x.cosa-cos3x.sina=1\)

\(\Leftrightarrow sin\left(3x-a\right)=1\)

\(\Leftrightarrow x=\frac{a}{3}+\frac{\pi}{6}+\frac{k2\pi}{3}\)