Đáp án:
$[_{x = \frac{\pi }{2} + k2\pi }^{x = k2\pi },k \in Z$
Giải thích các bước giải:
$\begin{array}{l}
4(\sin x + \cos x) - 3\sin x\cos x - 4 = 0(*)\\
{(\sin x + \cos x)^2} = {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x = 1 + 2\sin x\cos x\\
= > \sin x\cos x = \frac{{{{(\sin x + \cos x)}^2} - 1}}{2}\\
(*) = > 4(\sin x + \cos x) - 3\left( {\frac{{{{(\sin x + \cos x)}^2} - 1}}{2}} \right) - 4 = 0\\
< = > 4(\sin x + \cos x) - \frac{3}{2}{(\sin x + \cos x)^2} - \frac{5}{2} = 0\\
Đặt t = \sin x + \cos x = \sqrt 2 \sin (x + \frac{\pi }{4}) = > t \in [ - \sqrt 2 ;\sqrt 2 ]\\
PTTT:\\
- \frac{3}{2}{t^2} + 4t - \frac{5}{2} = 0\\
< = > [_{t = 1(nhận)}^{t = \frac{5}{3}(loại)}\\
= > \sqrt 2 \sin (x + \frac{\pi }{4}) = 1\\
< = > \sin (x + \frac{\pi }{4}) = \frac{{\sqrt 2 }}{2} = \sin \frac{\pi }{4}\\
< = > [_{x + \frac{\pi }{4} = \pi - \frac{\pi }{4} + k2\pi }^{x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi }\\
< = > [_{x = \frac{\pi }{2} + k2\pi }^{x = k2\pi },k \in Z
\end{array}$
