1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

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

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

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

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

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

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

a/

\(\Leftrightarrow sinx+cosx=\sqrt{2}sin2x\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}sin2x\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=sin2x\)

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

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

b/

\(\Leftrightarrow\frac{1-cos2x}{2}+sin2x=\frac{3\left(1+cos2x\right)}{2}\)

\(\Leftrightarrow sin2x-2cos2x=1\)

\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x-\frac{2}{\sqrt{5}}cos2x=\frac{1}{\sqrt{5}}\)

Đặt \(\frac{1}{\sqrt{5}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Leftrightarrow sin2x.cosa-cos2a.sina=cosa\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=a-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sinx-sin^2x=cosx-cos^2x\)

\(\Leftrightarrow sinx-cosx-\left(sin^2x-cos^2x\right)=0\)

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

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

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

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

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

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

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

d/

\(\Leftrightarrow2\left(sinx-cosx\right)\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(sinx-cosx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\2\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(2\right)\end{matrix}\right.\)

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

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\)

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

\(\left(2\right)\Leftrightarrow2+2sinx.cosx=\sqrt{3}cos2x\)

\(\Leftrightarrow2+sin2x=\sqrt{3}cos2x\)

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

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

\(\Leftrightarrow2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\frac{\pi}{12}+k\pi\)

    1 + sinx + cosx + sin2x + cos2x = 0

<=> sin^2x+ cos^2 x + ( sinx+cosx) + 2.sinx.cosx + ( cos^2 x - sin^2 x)=0

<=> 2 cos^2 x + 2sinx.cosx + sinx + cosx =0

<=> 2cosx ( cos x + sinx) + sinx + cosx = 0

<=> ( cosx + sinx ) (2 cos x + 1 ) = 0

<=> cosx + sinx = 0 hoặc 2cosx + 1 =0

1.

a, Phương trình có nghiệm khi: 

\(\left(m+2\right)^2+m^2\ge4\)

\(\Leftrightarrow m^2+4m+4+m^2\ge4\)

\(\Leftrightarrow2m^2+4m\ge0\)

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

b, Phương trình có nghiệm khi:

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

\(\Leftrightarrow2m^2+6m\le0\)

\(\Leftrightarrow-3\le m\le0\)

2.

a, Phương trình vô nghiệm khi:

\(\left(2m-1\right)^2+\left(m-1\right)^2< \left(m-3\right)^2\)

\(\Leftrightarrow4m^2-4m+1+m^2-2m+1< m^2-6m+9\)

\(\Leftrightarrow4m^2-7< 0\)

\(\Leftrightarrow-\dfrac{\sqrt{7}}{2}< m< \dfrac{\sqrt{7}}{2}\)

b, \(2sinx+cosx=m\left(sinx-2cosx+3\right)\)

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

 Phương trình vô nghiệm khi:

\(\left(m-2\right)^2+\left(2m+1\right)^2< 9m^2\)

\(\Leftrightarrow m^2-4m+4+4m^2+4m+1< 9m^2\)

\(\Leftrightarrow m^2-1>0\)

\(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)

1.

c, \(\left(m+2\right)sin2x+mcos^2x=m-2+msin^2x\)

\(\Leftrightarrow\left(m+2\right)sin2x+m\left(cos^2x-sin^2x\right)=m-2\)

\(\Leftrightarrow\left(m+2\right)sin2x+mcos2x=m-2\)

Phương trình vô nghiệm khi:

\(\left(m+2\right)^2+m^2< \left(m-2\right)^2\)

\(\Leftrightarrow m^2+4m+4+m^2< m^2-4m+4\)

\(\Leftrightarrow m^2+8m< 0\)

\(\Leftrightarrow-8\le m\le0\)

a/

\(\left(m+1\right)^2+\left(m-1\right)^2\ge\left(2m+3\right)^2\)

\(\Leftrightarrow2m^2+12m+7\le0\)

\(\Leftrightarrow\frac{-6-\sqrt{22}}{2}\le m\le\frac{-6+\sqrt{22}}{2}\)

b/ \(\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\\left(m-1\right)^2+4m\ge m^4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\m^4-\left(m+1\right)^2\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\\left(m^2+m+1\right)\left(m^2-m-1\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow0\le m\le\frac{1+\sqrt{5}}{2}\)

c/ \(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}=m\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)+\frac{1}{2}=m\)

Do \(-\frac{1}{2}\le sin\left(2x-\frac{\pi}{3}\right)\le\frac{3}{2}\Rightarrow-\frac{1}{2}\le m\le\frac{3}{2}\)