Question

\(4sin^22x+8cos^2x-\frac{19}{3}=0\)

Answer 1

Lời giải:

PT $\Leftrightarrow 4(2\sin x\cos x)^2+8\cos ^2x-\frac{19}{3}=0$

$\Leftrightarrow 16\cos ^2x(1-\cos ^2x)+8\cos ^2x-\frac{19}{3}=0$

$\Leftrightarrow -16\cos ^4x+24\cos ^2x-\frac{19}{3}=0$

$\Leftrightarrow -16a^2+24a-\frac{19}{3}=0$ (đặt $a=\cos ^2x$. ĐK: $a\in [0;1]$)

$\Rightarrow a=\frac{9\pm 2\sqrt{6}}{12}$

Do $a\in [0;1]$ nên $a=\cos ^2x=\frac{9-2\sqrt{6}}{12}$

$\Rightarrow \cos 2x=2\cos ^2x-1=\frac{3-2\sqrt{6}}{6}$

\(\Rightarrow x=k\pi\pm \frac{1}{2}\cos ^{-1}\frac{3-2\sqrt{6}}{6}\) với $k$ nguyên.