BÀi 1:
a: \(1-5\cdot\sin x+2\cdot cos^2x=0\)
=>\(2\left(1-\sin^2x\right)-5\cdot\sin x+1=0\)
=>\(2-2\cdot\sin^2x-5\cdot\sin x+1=0\)
=>\(-2\cdot\sin^2x-5\cdot\sin x+3=0\)
=>\(2\cdot\sin^2x+5\cdot\sin x-3=0\)
=>(sin x+3)(2sin x-1)=0
mà sin x+3>0
nên 2 sin x-1=0
=>sin x=1/2
=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)
b: \(cos2x+\sin^2x+2\cdot cosx+1=0\)
=>\(2\cdot cos^2x-1+1-cos^2x+2\cdot cosx+1=0\)
=>\(cos^2x+2\cdot cosx+1=0\)
=>\(\left(cosx+1\right)^2=0\)
=>cosx+1=0
=>cosx=-1
=>\(x=\pi+k2\pi\)
c: \(\sin x-\sqrt3\cdot cosx=2\cdot\sin3x\)
=>\(\frac12\cdot\sin x-\frac{\sqrt3}{2}\cdot cosx=\sin3x\)
=>\(\sin\left(x-\frac{\pi}{3}\right)=\sin3x\)
=>\(\left[\begin{array}{l}3x=x-\frac{\pi}{3}+k2\pi\\ 3x=\pi-x+\frac{\pi}{3}+k2\pi=\frac43\pi-x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=-\frac{\pi}{3}+k2\pi\\ 4x=\frac43\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=-\frac{\pi}{6}+k\pi\\ x=\frac13\pi+\frac{k\pi}{2}\end{array}\right.\)
d: \(cos3x-\sin x=\sqrt3\left(cosx-\sin3x\right)\)
=>\(cos3x+\sqrt3\cdot\sin3x=\sqrt3\cdot cosx+\sin x\)
=>\(\sin3x\cdot\frac{\sqrt3}{2}+cos3x\cdot\frac12=\sin x\cdot\frac12+\frac{\sqrt3}{2}\cdot cosx\)
=>\(\sin\left(3x+\frac{\pi}{6}\right)=\sin\left(x+\frac{\pi}{3}\right)\)
=>\(\left[\begin{array}{l}3x+\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\ 3x+\frac{\pi}{6}=\pi-x-\frac{\pi}{3}+k2\pi=\frac23\pi-x+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}-\frac{\pi}{6}+k2\pi=\frac{\pi}{6}+k2\pi\\ 4x=\frac23\pi-\frac{\pi}{6}+k2\pi=\frac{\pi}{2}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{12}+k\pi\\ x=\frac{\pi}{8}+\frac{k\pi}{2}\end{array}\right.\)
