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

1.

Đề là \(x\in\left(0;\frac{\pi}{4}\right)\) hay \(x\in\left[0;\frac{\pi}{4}\right]\) ?

2.

\(sin3x-4sinx.cos2x=0\)

\(\Leftrightarrow sin3x-\left(2sin3x-2sinx\right)=0\)

\(\Leftrightarrow2sinx-sin3x=0\)

\(\Leftrightarrow2sinx-3sinx+4sin^3x=0\)

\(\Leftrightarrow sinx\left(4sin^2x-1\right)=0\)

\(\Leftrightarrow sinx\left(1-2cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)

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

3.

\(sin^2x.cosx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

4.

\(\sqrt{3}sin2x+1-cos2x=3\)

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

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{3}+k\pi\)

5.

Ko có 4 đáp án thì làm sao biết, có vô số pt tương đương với pt này :)

6.

\(sinx+cosx-2sinx.cosx+1=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t+1-t^2+1=0\)

\(\Leftrightarrow-t^2+t+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2sinx.cosx=t^2-1=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

Có b nào gipus mk với cần gấp gấp :)

1.

\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

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

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

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

2.

\(\left|cosx-sinx\right|+2sin2x=1\)

\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)

\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)

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

3.

\(2sin2x-3\sqrt{6}\left|sinx+cosx\right|+8=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^2-3\sqrt{6}\left|sinx+cosx\right|+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|sinx+cosx\right|=\sqrt{6}\left(vn\right)\\\left|sinx+cosx\right|=\dfrac{\sqrt{6}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left|sin\left(x+\dfrac{\pi}{4}\right)\right|=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\pm\dfrac{\sqrt{3}}{2}\)

...

Hướng dẫn giải:

Chọn A.

Ta có: sin²x – 2( m- 1)sinx. cosx – (m- 1).cos²x = m

1.

Bạn xem lại đề, \(sin^2x\left(\frac{x}{2}-\frac{\pi}{4}\right)\) là sao nhỉ?Có cả x trong lẫn ngoài ngoặc?

2.

ĐKXĐ: \(sinx\ne0\)

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

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

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

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

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

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

3.

Theo điều kiện của pt lượng giác bậc nhất:

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

\(\Leftrightarrow10m^2+6m+1\ge4m^2-4m+1\)

\(\Leftrightarrow3m^2+5m\ge0\Rightarrow\left[{}\begin{matrix}m\ge0\\m\le-\frac{5}{3}\end{matrix}\right.\)

4.

\(\Leftrightarrow1-sin^2x-\left(m^2-3\right)sinx+2m^2-3=0\)

\(\Leftrightarrow-sin^2x-m^2sinx+2m^2+3sinx-2=0\)

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

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

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

\(\Leftrightarrow sinx=1-m^2\)

\(\Rightarrow-1\le1-m^2\le1\)

\(\Rightarrow m^2\le2\Rightarrow-\sqrt{2}\le m\le\sqrt{2}\)