Question

[Trần Khánh Huyền]

\(\sin^6+\cos^6x=4\cos^22x\)

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=4cos^22x\)

\(\Leftrightarrow1-3sin^2x.cos^2x=4\left(1-sin^22x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=4-4sin^22x\)

\(\Leftrightarrow sin^22x=\frac{12}{13}\)

\(\Leftrightarrow1-cos4x=\frac{24}{13}\)

\(\Leftrightarrow cos4x=-\frac{11}{13}\)

\(\Rightarrow x=\pm\frac{1}{4}arccos\left(-\frac{11}{13}\right)+k2\pi\)