\(y=1-2\left(2sinx.cosx\right)^2+2sin^42x=1-2sin^22x+2sin^42x\)

\(y=1-2sin^22x\left(1-sin^22x\right)=1-2sin^22x.cos^22x\)

\(=1-\dfrac{1}{2}\left(2sin2x.cos2x\right)^2=1-\dfrac{1}{2}sin^24x\)

Do \(0\le sin^24x\le1\Rightarrow\dfrac{1}{2}\le y\le1\)

\(y_{min}=\dfrac{1}{2}\) khi \(sin^24x=1\)

\(y_{max}=1\) khi \(sin4x=0\)

46.

\(cos2x+\left(2m+1\right)cosx+2m=0\)

\(\Leftrightarrow2cos^2x-1+\left(2m+1\right)cosx+2m=0\)

\(\Leftrightarrow2cos^2x+cosx-1+2m\left(cosx+1\right)=0\)

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

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

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

Do (1) có đúng 1 nghiệm \(x=\pi\) thuộc khoảng đã cho nên pt đã cho có nghiệm duy nhất thuộc \([0;2\pi)\) khi và chỉ khi:

\(\left[{}\begin{matrix}\dfrac{2m-1}{2}=-1\\\dfrac{2m-1}{2}>1\\\dfrac{2m-1}{2}< -1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m>\dfrac{3}{2}\\m\le-\dfrac{1}{2}\end{matrix}\right.\)

47.

\(\sqrt{3}sinx+cosx=m\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{m}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{m}{2}\)

Đặt \(x+\dfrac{\pi}{6}=t\Rightarrow t\in\left[0;3\pi\right]\)

Từ đường tròn lượng giác ta thấy \(sint=\dfrac{m}{2}\) trên \(\left[0;3\pi\right]\):

- Có 1 nghiệm khi \(\dfrac{m}{2}=-1\)

- Có 2 nghiệm khi \(-1< \dfrac{m}{2}< 0\) hoặc \(\dfrac{m}{2}=1\)

- Có 4 nghiệm khi \(0\le\dfrac{m}{2}< 1\)

\(\Rightarrow\) Không tồn tại m để pt đã cho có đúng 3 nghiệm trên miền đã cho 

Cả 4 đáp án đều sai

Với \(m=0\) pt vô nghiệm

Với \(m\ne0\Rightarrow cot^2\left(2x-\dfrac{\pi}{8}\right)=\dfrac{2m+1}{m}\)

Do \(cot^2\left(2x-\dfrac{\pi}{8}\right)\ge0\Rightarrow\) pt có nghiệm khi:

\(\dfrac{2m+1}{m}\ge0\Rightarrow\left[{}\begin{matrix}m>0\\m\le-\dfrac{1}{2}\end{matrix}\right.\)

Tọa độ G là:

\(\left\{{}\begin{matrix}x_G=\dfrac{1+0+2}{3}=1\\y_G=\dfrac{0+2+4}{3}=2\end{matrix}\right.\)

Tọa độ G' là:

\(\left\{{}\begin{matrix}x_{G'}=1+2=3\\y_{G'}=2-1=1\end{matrix}\right.\)

7.

\(3cos^2x=-8cosx-5\)

\(\Leftrightarrow3cos^2x+8cosx+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\cosx=-\dfrac{5}{3}< -1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=\pi+k2\pi\)

8.

\(2sin^2x-3sinx+1=0\)

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

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

\(\Rightarrow x=\dfrac{\pi}{6}\)

9.

\(sin^2x-4sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=4>1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=k\pi\)

10.

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{3}{2}< -1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow2cos^2x-\left(2m+1\right)cosx+m=0\)

\(\Leftrightarrow2cos^2x-2mcosx-cosx+m=0\)

\(\Leftrightarrow2cosx\left(cosx-m\right)-\left(cosx-m\right)=0\)

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

Do \(2cosx-1=0\) ko có nghiệm trên đoạn đã cho nên \(cosx-m=0\) có nghiệm trên \(\left[\dfrac{\pi}{2};\pi\right]\)

\(x\in\left[\dfrac{\pi}{2};\pi\right]\Rightarrow cosx\in\left[-1;0\right]\)

\(\Rightarrow-1\le m\le0\)

\(cos2x+cos^2x+3sinx+2m=0\)

\(\Leftrightarrow1-2sin^2x+1-sin^2x+3sinx+2m=0\)

\(\Leftrightarrow3sin^2x-3sinx-2=2m\)

Đặt \(sinx=t\in\left[-1;1\right]\)

\(\Rightarrow3t^2-3t-2=2m\)

Xét hàm \(f\left(t\right)=3t^2-3t-2\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=\dfrac{1}{2}\in\left[-1;1\right]\)

\(f\left(-1\right)=4\) ; \(f\left(\dfrac{1}{2}\right)=-\dfrac{11}{4}\) ; \(f\left(1\right)=-2\)

\(\Rightarrow-\dfrac{11}{4}\le f\left(t\right)\le4\)

\(\Rightarrow\) Pt đã cho có nghiệm khi \(-\dfrac{11}{4}\le2m\le4\Rightarrow-\dfrac{11}{8}\le m\le2\)

17.

\(cosx.cos3x=-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x=-\dfrac{1}{2}\)

\(\Leftrightarrow2cos^22x-1+cos2x=-1\)

\(\Leftrightarrow2cos^22x-cos2x=0\)

\(\Rightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

17: \(\Leftrightarrow\dfrac{1}{2}\cdot\left(cos\left(-2x\right)+cos4x\right)=\dfrac{-1}{2}\)

=>\(cos2x+cos4x=-1\)

=>\(2cos^22x-1+cos2x=-1\)

=>cos2x(2 cos 2x+1)=0

=>cos2x=0 hoặc cos2x=-1/2

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{pi}{2}+kpi\\2x=\dfrac{2}{3}pi+k2pi\\2x=-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{pi}{4}+\dfrac{kpi}{2}\\x=\dfrac{1}{3}pi+kpi\\x=-\dfrac{1}{3}pi+kpi\end{matrix}\right.\)

19: \(\Leftrightarrow4-4cos^2x+\left(2\sqrt{3}+2\right)cosx-4-\sqrt{3}=0\)

\(\Leftrightarrow4cos^2x-\left(2\sqrt{3}+2\right)cosx+\sqrt{3}=0\)

\(\text{Δ}=16-8\sqrt{3}\)

Pt sẽ có hai nghiệm là:

\(\left\{{}\begin{matrix}cosx=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{8}=\dfrac{4}{8}=\dfrac{1}{2}\\cosx=\dfrac{2\sqrt{3}+2+2\sqrt{3}-2}{8}=\dfrac{4\sqrt{3}}{8}=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{pi}{6}+k2pi\\x=\pm\dfrac{pi}{3}+k2pi\end{matrix}\right.\)