Soit \(x\in\Rr\).

\[\begin{align*} 2^{4\cos^2x+1}+16.2^{4\sin^2x-3}=20&\Leftrightarrow2^{4\cos^2x+1}+16.2^{1-4\cos^2x}=20\Leftrightarrow2^{4\cos^2x}-10+16\times2^{-4\cos^2x}=0\\ &\Leftrightarrow2^{4\cos^2x}-10+\frac{16}{2^{4\cos^2x}}=0\Leftrightarrow(2^{4\cos^2x})^2-10\times2^{4\cos^2x}+16=0\\ &\Leftrightarrow2^{4\cos^2x}=2\;\mbox{ou}\;2^{4\cos^2x}=8\Leftrightarrow4\cos^2x=1\;\mbox{ou}\;4\cos^2x=3\\ &\Leftrightarrow\cos x=\frac{1}{2}\;\mbox{ou}\;\cos x=-\frac{1}{2}\;\mbox{ou}\;\cos x=\frac{\sqrt{3}}{2}\;\mbox{ou}\;\cos x=-\frac{\sqrt{3}}{2}\\ &\Leftrightarrow x\in\left(\frac{\pi}{6}+\frac{\pi}{2}\Zz\right)\cup\left(\frac{\pi}{3}+\frac{\pi}{2}\Zz\right). \end{align*}\]

exo7 5071