a.

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)

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

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

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

c.

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

a/ Bạn coi lại vế trái đề bài, nhìn không hợp lý

b/ \(\Leftrightarrow\frac{1}{2}sin9x-\frac{1}{2}sinx=\frac{1}{2}sin5x-\frac{1}{2}sinx\)

\(\Leftrightarrow sin9x=sin5x\)

\(\Leftrightarrow\left[{}\begin{matrix}9x=5x+k2\pi\\9x=\pi-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{14}+\frac{k\pi}{7}\end{matrix}\right.\)

c/ \(\Leftrightarrow sin2x-cos2x=cosx-sinx\)

\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

\(\Leftrightarrow cos\left(\frac{3\pi}{4}-2x\right)=cos\left(x+\frac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{3\pi}{4}-2x=x+\frac{\pi}{4}+k2\pi\\\frac{3\pi}{4}-2x=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\pi+k2\pi\end{matrix}\right.\)

d/

\(\Leftrightarrow sin2x=sin6x-sin4x\)

\(\Leftrightarrow2sinx.cosx=2cos5x.sinx\)

\(\Leftrightarrow sinx\left(cosx-cos5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos5x=cosx\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\5x=x+k2\pi\\5x=-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\)

a/

\(\sqrt{3}sin4x-cos4x=2cosx\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=cosx\)

\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{6}=\frac{\pi}{2}-x+k2\pi\\4x-\frac{\pi}{6}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)

b/

\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)

\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)

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

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)

\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

Lời giải:
PT $\Leftrightarrow (2\cos x-1)(2\sin x+\cos x)=2\sin x\cos x-\sin x$

$\Leftrightarrow (2\cos x-1)(2\sin x+\cos x)=\sin x(2\cos x-1)$

$\Leftrightarrow (2\cos x-1)(\sin x+\cos x)=0$

$\Rightarrow 2\cos x=1$ hoặc $\sin x=-\cos x=\cos (\pi -x)=\sin (x-\frac{\pi}{2})$

Đến đây thì đơn giản rồi.

\(\dfrac{cosx+sin2x}{cos2x+sinx}=\sqrt{3}\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x+\dfrac{\pi}{6}\right)\)

Làm nốt nhé

a/

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{3}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\sqrt{\frac{3}{2}}>1\)

Pt vô nghiệm

b/

\(\Leftrightarrow\frac{2}{\sqrt{13}}sinx+\frac{3}{\sqrt{13}}cosx=\frac{2}{\sqrt{13}}\)

Đặt \(\frac{2}{\sqrt{13}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow sinx.cosa+cosx.sina=cosa\)

\(\Leftrightarrow sin\left(x+a\right)=sin\left(\frac{\pi}{2}-a\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+a=\frac{\pi}{2}-a+k2\pi\\x+a=\frac{\pi}{2}+a+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}-2a+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=x+\frac{\pi}{6}+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

2.

Theo điều kiện có nghiệm của pt lượng giác bậc nhất với sin và cos:

\(m^2+\left(m-1\right)^2\ge5\)

\(\Leftrightarrow m^2-m-2\ge0\Leftrightarrow\left[{}\begin{matrix}m\ge2\\m\le-1\end{matrix}\right.\)

Pt <=> 2sin\(\dfrac{3x}{2}\).cos\(\dfrac{x}{2}\) = 2cos\(\dfrac{3x}{2}\).cos\(\dfrac{x}{2}\)

⇔ cos\(\dfrac{x}{2}\) . \(\left(sin\dfrac{3x}{2}-cos\dfrac{3x}{2}\right)\) = 0

⇔ \(\sqrt{2}sin\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right).cos\dfrac{x}{2}=0\)

⇔