Question

giải phương trình \(\sqrt[4]{10+8\cos^2x}-\sqrt[8]{8\sin^2x-1}=1\)

Answer 1

\(\sqrt[4]{10+8\left(1-sin^2x\right)}-\sqrt[4]{8sin^2x-1}=1\)

\(\Leftrightarrow\sqrt[4]{18-8sin^2x}-\sqrt[4]{8sin^2x-1}=1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[4]{18-8sin^2x}=a>0\\\sqrt[4]{8sin^2x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^4+b^4=17\end{matrix}\right.\) \(\Rightarrow\left(b+1\right)^4+b^4-17=0\)

\(\Leftrightarrow\left(b-1\right)\left(b+2\right)\left(b^2+b+4\right)=0\)

\(\Leftrightarrow b=1\Leftrightarrow\sqrt[4]{8sin^2x-1}=1\)

\(\Leftrightarrow4sin^2x-1=0\Leftrightarrow2\left(1-cos2x\right)-1=0\)

\(\Leftrightarrow cos2x=\frac{1}{2}\Leftrightarrow x=...\)