\(x\ne\left\{k\pi\right\}\)
Pt \(\Leftrightarrow\)\(5cosx-3\left(1-cosx\right).\dfrac{cos^2x}{1-cos^2x}=2\)
Đặt \(t=cosx,t\in\left(-1;1\right)\)
Pttt:\(5t-\dfrac{3\left(1-t\right)t^2}{1-t^2}=2\)
\(\Leftrightarrow5t-\dfrac{3t^2}{1+t}=2\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}t=\dfrac{1}{2}\left(tm\right)\\t=-2\left(vn\right)\end{matrix}\right.\)\(\Rightarrow cosx=\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
Vậy...
`5 sin ([5\pi]/2-x)-3(1-cos x).cot^2 x=2` `ĐK: x \ne k\pi` `(k in ZZ)`
`<=>5 sin(\pi/2-x)-3(1-cos x).[cos^2 x]/[sin^2 x]=2`
`<=>5 cos x-(3-3cos x).[cos^2 x]/[sin^2 x]=2`
`<=>[5 cos x . sin^2 x-3cos^2 x+3 cos^3 x]/[sin^2 x]=[2sin^2 x]/[sin^2 x]`
`=>5cos x.(1-cos^2 x)-3cos^2 x+3 cos^3 x=2(1-cos^2 x)`
`<=>5cos x -5cos^3 x-3cos^2 x+3 cos^3 x=2-2cos^2 x`
`<=>2cos^3 x+cos^2 x-5 cos x+2=0`
`<=>` $\left[\begin{matrix} cos x=1\\ cos x=\dfrac{1}{2}\\cos x=-2(VN)\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x=k2\pi (ko t/m)\\ x=\dfrac{+-\pi}{3}+k2\pi (t/m)\end{matrix}\right.$ `(k in ZZ)`
`<=>x=[+-\pi]/3+k2\pi` `(k in ZZ)`
