Giải phương trình lượng giác sau: \({\sin}^6 x+{\cos}^6+\dfrac{1}{2}\sin 4x=0\)

Ta có \({\sin}^6 x+{\cos}^6 x+\dfrac{1}{2}\sin 4x=0\)

\(\Leftrightarrow ({{\sin}^2 x+{\cos}^2 x)}^3\)

\(-3{\sin}^2 x{\cos}^2 x({\sin}^2 x+{\cos}^2 x)\)

\(+\dfrac{1}{2}\sin 4x=0\)

\(\Leftrightarrow 1-3{\sin}^2 x{\cos}^2 x+\dfrac{1}{2}\sin 4x=0\)

\(\Leftrightarrow 1-3{\left({\dfrac{\sin 2x}{2}}\right)}^2+\dfrac{1}{2}\sin 4x=0\)

\(\Leftrightarrow 1-\dfrac{3}{4}{\sin}^2 2x+\dfrac{1}{2}\sin 4x=0\)

\(\Leftrightarrow 1-\dfrac{3}{4}\dfrac{1-\cos 4x}{2}+\dfrac{1}{2}\sin 4x=0\)

\(\Leftrightarrow 8-3+3\cos 4x+4\sin 4x=0\)

\(\Leftrightarrow 3\cos 4x+4\sin 4x=-5\)

\(\Leftrightarrow \dfrac{3}{5}\cos 4x+\dfrac{4}{5}\sin 4x=-1\)

Đặt \(\dfrac{3}{5}=\sin\alpha\), \(\dfrac{4}{5}=\cos\alpha\) ta được

\(\sin\alpha\cos 4x+\cos \alpha\sin 4x=-1\)

\(\Leftrightarrow \sin (4x+\alpha)=-1\)

\(\Leftrightarrow 4x+\alpha=\dfrac{3\pi}{2}+k2\pi,k\in\mathbb{Z}\)

\(\Leftrightarrow x=\dfrac{3\pi}{8}-\dfrac{\alpha}{4}+k\dfrac{\pi}{2},k\in\mathbb{Z}\)

Vậu phương trình có nghiệm là \(x=\dfrac{3\pi}{8}-\dfrac{\alpha}{4}+k\dfrac{\pi}{2},k\in\mathbb{Z}\).