ĐKXĐ: \(x<>\frac{\pi}{2}+k\pi\)

\(\sqrt3\cdot\cot^2x-4\cdot\cot x+\sqrt3=0\) (1)

\(\Delta=\left(-4\right)^2-2\cdot\sqrt3\cdot\sqrt3=16-2\cdot3=16-6=10>0\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}\cot x=\frac{4-\sqrt{10}}{2\cdot\sqrt3}=\frac{4-\sqrt{10}}{\sqrt{12}}=\frac{4\sqrt3-\sqrt{30}}{6}\\ cotx=\frac{4+\sqrt{10}}{2\cdot\sqrt3}=\frac{4+\sqrt{10}}{\sqrt{12}}=\frac{4\sqrt3+\sqrt{30}}{6}\end{array}\right.\)

TH1: \(\cot x=\frac{4\sqrt3-\sqrt{30}}{6}\)

=>\(x=arc\cot\left(\frac{4\sqrt3-\sqrt{30}}{6}\right)+k\pi\) (nhận)

TH2: \(\cot x=\frac{4\sqrt3+\sqrt{30}}{6}\)

=>\(x=arc\cot\left(\frac{4\sqrt3+\sqrt{30}}{6}\right)+k\pi\) (nhận)

1: \(2\cdot cos^2x+7\cdot\sin x-5=0\)

=>\(2\left(1-\sin^2x\right)+7\cdot\sin x-5=0\)

=>\(2-2\cdot\sin^2x+7\cdot\sin x-5=0\)

=>\(-2\cdot\sin^2x+7\cdot\sin x-3=0\)

=>\(2\cdot\sin^2x-7\cdot\sin x+3=0\)

=>(sin x-3)(2sin x-1)=0

TH1: sin x-3=0

=>sin x=3(vô lý)

=>Loại

TH2: 2 sin x-1=0

=>2 sin x=1

=>\(\sin x=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)

2: \(2\cdot cos^2x+5\cdot\sin x-4=0\)

=>\(2\left(1-\sin^2x\right)+5\cdot\sin x-4=0\)

=>\(-2\cdot\sin^2x+5\cdot sinx-2=0\)

=>\(-\left(2\cdot\sin x-1\right)\left(\sin x-2\right)=0\)

TH1: sin x-2=0

=>sin x=2

=>Loại

TH2: 2 sin x-1=0

=>2 sin x=1

=>\(\sin x=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)

3: cos2x+3sin x=2

=>\(1-2\cdot\sin^2x+3\cdot\sin x-2=0\)

=>\(-2\cdot\sin^2x+3\cdot\sin x-1=0\)

=>\(2\cdot\sin^2x-3\cdot\sin x+1=0\)

=>(sin x-1)(2sin x-1)=0

TH1: sin x-1=0

=>sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

TH2: 2 sin x-1=0

=>2 sin x=1

=>\(\sin x=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)

4: cos2x+3cosx-4=0
=>\(2\cdot cos^2x-1+3\cdot cosx-4=0\)

=>\(2\cdot cos^2x+3\cdot cosx-5=0\)

=>\(2\cdot cos^2x+5\cdot cos^2x-2\cdot cosx-5=0\)

=>(2cosx-5)(cosx-1)=0

TH1: 2 cosx-5=0

=>cosx=5/2

=>Loại

Th2: cos x-1=0

=>cosx=1

=>\(x=k2\pi\)

5: \(cosx+cos\left(\frac{x}{2}\right)+1=0\)

=>\(2\cdot cos^2\left(\frac{x}{2}\right)-1+cos\left(\frac{x}{2}\right)+1=0\)

=>\(2\cdot cos^2\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right)=0\)

=>\(cos\left(\frac{x}{2}\right)\cdot\left\lbrack2\cdot cos^{}\left(\frac{x}{2}\right)+1\right\rbrack=0\)

TH1: \(cos\left(\frac{x}{2}\right)=0\)

=>\(\frac{x}{2}=\frac{\pi}{2}+k\pi\)

=>\(x=\pi+k2\pi\)

TH2: \(2\cdot cos\left(\frac{x}{2}\right)+1=0\)

=>\(cos\left(\frac{x}{2}\right)=-\frac12\)

=>\(\left[\begin{array}{l}\frac{x}{2}=\frac23\pi+k2\pi\\ \frac{x}{2}=-\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\pi+k4\pi\\ x=-\frac43\pi+k4\pi\end{array}\right.\)

\(\Leftrightarrow\left(sinx-3\right)\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=3>1\left(loại\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)