a,
$sin5x+cos5x=-\sqrt2$
$\Leftrightarrow \frac{1}{\sqrt2}sin5x+ \frac{1}{\sqrt2}cos5x=-1$
$\Leftrightarrow sin(5x+\frac{\pi}{4})=-1$
- TH1: $5x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi$
$\Leftrightarrow x=\frac{-3\pi}{20}+\frac{k2\pi}{5}$
- TH2: $5x+\frac{\pi}{4}= \frac{3\pi}{2}+k2\pi$
$\Leftrightarrow x=\frac{\pi}{4}+\frac{k2\pi}{5}$
b,
$3sin^2x+2sinx.cosx+cos^2x=3$
1. $cos^2x=0$
$\Rightarrow sin^2x=1$ (TM)
2. $cos^2x\neq 0$
$\Rightarrow 3tan^2x+2tanx+1=\frac{3}{cos^2x}$
$\Leftrightarrow 3tan^2x+2tanx+1=3(1+tan^2x)$
$\Leftrightarrow 2tanx=2$
$\Leftrightarrow tanx=1$
$\Leftrightarrow x=\frac{\pi}{4}+k\pi$
c, (Giống a, chia 2 vế cho 2)
