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a: =>sin2x+2*(1-cos2x)/2=2

=>sin2x-cos2x=1

=>căn 2*sin(2x-pi/4)=1

=>2x-pi/4=pi/4+k2pi hoặc 2x-pi/4=3/4pi+k2pi

=>x=pi/4+kpi hoặc x=pi/2+kpi

b: =>2*(1+cos2x)/2+1/2*sin2x-1/2(1-cos2x)=0

=>1+cos2x+1/2*sin2x-1/2+1/2cos2x=0

=>1/2*sin2x+3/2*cos2x=-1/2

=>sin(2x+a)=-cos(a)=cos(pi-a)

=>sin(2x+a)=sin(-pi/2+a)

=>2x+a=-pi/2+a+k2pi hoặc 2x+a=3/2pi-a+k2pi

=>x=-pi/4+kpi hoặc x=3/4pi-a+kpi

a/

\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=-\frac{1}{3}\end{matrix}\right.\) (đặt \(cosx=t\) thành pt bậc 2 rồi bấm máy ra nghiệm thôi)

\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm arccos\left(-\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)

b/

\(\Leftrightarrow6\left(1-sin^2x\right)+5sinx-7=0\)

\(\Leftrightarrow-6sin^2x+5sinx-1=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{5}{2}\left(l\right)\end{matrix}\right.\)

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

d/

\(\Leftrightarrow2cos^2\frac{x}{2}-1+3cos\frac{x}{2}+2=0\)

\(\Leftrightarrow2cos^2\frac{x}{2}+3cos\frac{x}{2}+1=0\)

\(\Rightarrow\left[{}\begin{matrix}cos\frac{x}{2}=-1\\cos\frac{x}{2}=-\frac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\pi+k4\pi\\x=\pm\frac{4\pi}{3}+k4\pi\end{matrix}\right.\)

\(y=2sin^2x+3sinx.cosx+cos^2x\)

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

\(=-cos2x+\dfrac{3}{2}sin2x+\dfrac{1}{2}cos2x+\dfrac{1}{2}\)

\(=\dfrac{3}{2}sin2x-\dfrac{1}{2}cos2x+\dfrac{1}{2}\)

\(=\dfrac{\sqrt{10}}{2}\left(\dfrac{3}{\sqrt{10}}sin2x-\dfrac{1}{\sqrt{10}}cos2x\right)+\dfrac{1}{2}\)

\(=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\)

Vì \(sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)\in\left[-1;1\right]\)

\(\Rightarrow y=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\in\left[-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2};\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\right]\)

\(\Rightarrow y_{min}=-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=-1\Leftrightarrow...\)

\(y_{max}=\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=1\Leftrightarrow...\)

\(B=cos^2x.cot^2x+cos^2x-cot^2x+2\left(sin^2x+cos^2x\right)\)

\(=cos^2x\left(cot^2x+1\right)-cot^2x+2\)

\(=\frac{cos^2x}{sin^2x}-cot^2x+1=cot^2x-cot^2x+1=1\)

\(M=cos^4x-sin^4x+cos^4x+sin^2x.cos^2x+3sin^2x\)

\(=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)

\(=cos^2x-sin^2x+cos^2x+3sin^2x\)

\(=2\left(sin^2x+cos^2x\right)=2\)

a/ \(2\left(1-cos^2x\right)+3cos^2x-2=m\)

\(\Leftrightarrow cos^2x=m\)

Do \(0\le cos^2x\le1\) nên pt có nghiệm khi và chỉ khi \(0\le m\le1\)

b/ \(\Leftrightarrow\left\{{}\begin{matrix}cosx=m\\sinx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}cosx=m\\cosx\ne\pm1\end{matrix}\right.\)

\(\Rightarrow-1< m< 1\)

Chọn B

\(2sin2x+sinx.cosx-cos^2x+1=0\)

\(\Leftrightarrow4sin2x+2sinx.cosx-2cos^2x+2=0\)

\(\Leftrightarrow4sin2x+sin2x-cos2x=-1\)

\(\Leftrightarrow5sin2x-cos2x=-1\)

\(\Leftrightarrow\sqrt{26}\left(\dfrac{5}{\sqrt{26}}sin2x-\dfrac{1}{\sqrt{26}}cos2x\right)=-1\)

\(\Leftrightarrow cos\left(2x+arccos\dfrac{1}{\sqrt{26}}\right)=\dfrac{1}{\sqrt{26}}\)

\(\Leftrightarrow2x+arccos\dfrac{1}{\sqrt{26}}=\pm arccos\dfrac{1}{\sqrt{26}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-arccos\dfrac{1}{\sqrt{26}}+k\pi\end{matrix}\right.\)