b:

ĐKXĐ: \(\left\{{}\begin{matrix}cosx< >0\\sinx< >0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{2}+k\Omega\\x\ne k\Omega\end{matrix}\right.\)

=>\(x\ne\dfrac{\Omega}{2}+\dfrac{k\Omega}{2}\)

 \(\dfrac{1}{cosx}+\dfrac{\sqrt{3}}{sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left[sinx\cdot\cos\dfrac{\Omega}{3}+sin\left(\dfrac{\Omega}{3}\right)\cdot cosx\right]\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left(\dfrac{1}{2}\cdot sinx+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\)

=>\(\left(sinx+\sqrt{3}\cdot cosx\right)\left(\dfrac{1}{cosx\cdot sinx}-1\right)=0\)

=>\(2\cdot\left(sinx\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\cdot\left(\dfrac{2}{2\cdot sinx\cdot cosx}-1\right)=0\)

=>\(2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\cdot\left(\dfrac{2}{sin2x}-1\right)=0\)

=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{\Omega}{3}\right)=0\\\dfrac{2}{sin2x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=k\Omega\\sin2x=2\left(loại\right)\end{matrix}\right.\)

=>\(x=-\dfrac{\Omega}{3}+k\Omega\)

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.

Đáp án D

a/

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

\(\Leftrightarrow sin\left(\frac{\pi}{3}-x\right)=sin4x\)

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

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

b/

\(\Leftrightarrow2sinx.cosx+4sinx.cos^2x-2sinx=0\)

\(\Leftrightarrow2sinx\left(cosx+2cos^2x-1\right)=0\)

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

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

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

c/

\(\Leftrightarrow2cos4x.sin3x=cos4x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2sin3x=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\3x=\frac{\pi}{6}+k2\pi\\3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

d/

\(\Leftrightarrow6sinx+3cosx+3=sinx-2cosx+3\)

\(\Leftrightarrow sinx+cosx=0\)

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

Chữ "s" nó dư á mọi người, tại em sửa không được :<

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:

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

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

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

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