a:
ĐKXĐ: sin x>=0 và cos x>=0
\(\sqrt{\sin x}=\sqrt{cosx}\)
=>sin x=cosx
=>\(\sin x-cosx=0\)
=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)=0\)
=>\(\sin\left(x-\frac{\pi}{4}\right)=0\)
=>\(x-\frac{\pi}{4}=k\pi\)
=>\(x=\frac{\pi}{4}+k\pi\)
mà sin x>=0 và cosx>=0
nên \(\begin{cases}x=\frac{\pi}{4}+k\pi\\ k\ge0\end{cases}\)
c:
ĐKXĐ: \(x<>\frac{\pi}{2}+k\pi\)
\(1+\sin x+cosx+\tan x=0\)
=>\(1+\sin x+cosx+\frac{\sin x}{cosx}=0\)
=>\(\frac{\sin x+cos^2x+\sin x\cdot cosx+cosx}{cosx}=0\)
=>\(\sin x\left(cosx+1\right)+cosx\left(cosx+1\right)=0\)
=>(sin x+cosx)(cosx+1)=0
TH1: sin x+cosx=0
=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=0\)
=>\(\sin\left(x+\frac{\pi}{4}\right)=0\)
=>\(x+\frac{\pi}{4}=k\pi\)
=>\(x=-\frac{\pi}{4}+k\pi\)
TH2: cosx+1=0
=>cosx=-1
=>\(x=\pi+k2\pi\)
