a: ĐKXĐ: sin 2x<>1

=>2x<>pi/2+k2pi

=>x<>pi/4+kpi

\(\dfrac{cos2x}{sin2x-1}=0\)

=>cos2x=0

=>2x=pi/2+kpi

=>x=pi/4+kpi/2

Kết hợp ĐKXĐ, ta được:

x=3/4pi+k2pi hoặc x=7/4pi+k2pi

b: cos(sinx)=1

=>sin x=kpi

=>sin x=0

=>x=kpi

c: \(2\cdot sin^2x-1+cos3x=0\)

=>cos3x+cos2x=0

=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)

=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)

=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi

=>x=-pi+k2pi hoặc x=pi/5+k2pi/5

e: cos3x=-cos7x

=>cos3x=cos(pi-7x)

=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi

=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2

\(\Leftrightarrow cos2x.cosx+2cos2x+sin2x.cosx-sinx=0\)

\(\Leftrightarrow cos2x.cosx+2cos2x+2sinx.cos^2x-sinx=0\)

\(\Leftrightarrow cos2x.cosx+2cos2x+sinx\left(2cos^2x-1\right)=0\)

\(\Leftrightarrow cos2x.cosx+2cos2x+sinx.cos2x=0\)

\(\Leftrightarrow cos2x\left(cosx+2+sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(2sinx.cosx-cosx-\left(1-2sin^2x\right)+3sinx-1=0\)

\(\Leftrightarrow cosx\left(2sinx-1\right)+2sin^2x+3sinx-2=0\)

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

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

\(\Leftrightarrow...\)