Phương trình đã cho tương đương với :

\(1+\frac{\sqrt{3}}{2}\sin2x-\frac{1}{2}\cos2x-3\left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right)=0\)

\(\Leftrightarrow1-\cos\left(2x+\frac{\pi}{3}\right)-3\sin\left(x+\frac{\pi}{6}\right)=0\)

\(2\sin^2\left(x+\frac{\pi}{6}\right)-2\sin\left(x+\frac{\pi}{6}\right)=0\Leftrightarrow\begin{cases}\sin\left(x+\frac{\pi}{6}\right)=0\\\sin\left(x+\frac{\pi}{6}\right)=\frac{3}{2}\end{cases}\) (Loại \(\sin\left(x+\frac{\pi}{6}\right)=\frac{3}{2}\))

Với \(\sin\left(x+\frac{\pi}{6}\right)=0\Rightarrow x=-\frac{\pi}{6}+k\pi,k\in Z\)

A

A

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

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

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

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

\(\Leftrightarrow...\)

a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)

\(\Leftrightarrow3cos4x-cos6x-2=0\)

Đặt \(t=2x\)

Pttt:\(3cos2t-cos3t-2=0\)

\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)

\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))

\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)

\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)

\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))

Vậy...

a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)

Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)

\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)

\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

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

Vậy...

a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)

\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)

\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))

\(\Rightarrow sinx=1\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)

Vậy...

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

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

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

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

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

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

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

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