\(\Leftrightarrow2-6sinx.cosx-2sinx+2cosx+2cos^2x=0\)

\(\Leftrightarrow3\left(1-2sinx.cosx\right)-2\left(sinx-cosx\right)+cos^2x-sin^2x=0\)

\(\Leftrightarrow3\left(sinx-cosx\right)^2-2\left(sinx-cosx\right)-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)

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

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

Xét (1) \(\Leftrightarrow\frac{1}{\sqrt{5}}sinx-\frac{2}{\sqrt{5}}cosx=\frac{1}{\sqrt{5}}\)

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

\(\Rightarrow sinx.cosa-cosx.sina=cosa\)

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

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

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

a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)

\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)

Do \(-1\le sin\left(2x+\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le1\)

b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)

\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)

\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)

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

\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)

\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)

1.

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

Pt trở thành:

\(t^3+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow2t^3+t^2-3=0\)

\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)

\(\Leftrightarrow t=1\)

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

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

\(\Leftrightarrow...\)

b.

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t^4-3\left(t^2-1\right)-1=0\)

\(\Leftrightarrow t^4-3t^2+2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}1+sin2x=1\\1+sin2x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)+2\left(sinx+cosx\right)-6sinx.cosx=0\)

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

Pt trở thành:

\(t\left(1-\frac{t^2-1}{2}\right)+2t-3\left(t^2-1\right)=0\)

\(\Leftrightarrow-t^3-6t^2+7t+6=0\)

Nghiệm của pt bậc 3 này rất xấu, chắc bạn ghi ko đúng đề bài

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.

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

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

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

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

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

...

1B

2A

3A

4C

a.

\(\Leftrightarrow cosx\left[1-\left(1-2sin^2x\right)\right]-sin^2x=0\)

\(\Leftrightarrow2sin^2x.cosx-sin^2x=0\)

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

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

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

b.

Câu b chắc chắn đề đúng chứ bạn? Vế phải ấy?

c/

\(\left(1+cosx\right)\left(sinx-cosx+3\right)=1-cos^2x\)

\(\Leftrightarrow\left(1+cosx\right)\left(sinx-cosx+3\right)-\left(1+cosx\right)\left(1-cosx\right)=0\)

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

\(\Leftrightarrow cosx=-1\)

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

d.

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

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

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

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

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

a/

\(\Leftrightarrow sinx+cosx-4sinx.cosx-1=0\)

Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)

\(\Rightarrow t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)

Pt trở thành:

\(t-2\left(t^2-1\right)-1=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\)

b/

Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)

Pt trở thành:

\(t+\frac{3}{2}\left(t^2-1\right)-1=0\)

\(\Leftrightarrow3t^2+2t-5=0\)

\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=\frac{5}{3}>\sqrt{2}\left(l\right)\end{matrix}\right.\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

c/

\(\Leftrightarrow sinx+cosx-4sinx.cosx=\frac{1}{2}\)

Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\) với \(\left|t\right|\le\sqrt{2}\)

\(sinx.cosx=\frac{t^2-1}{2}\)

Pt trở thành:

\(t-2\left(t^2-1\right)=\frac{1}{2}\)

\(\Leftrightarrow-4t^2+2t+3=0\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{1+\sqrt{13}}{4}\\t=\frac{1-\sqrt{13}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{1+\sqrt{13}}{4\sqrt{2}}\\sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{13}}{4\sqrt{2}}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=arcsin\left(\frac{1+\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(\frac{1+\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=arcsin\left(\frac{1-\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(\frac{1-\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=...\)

1.

\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}+\sqrt{3}cos\frac{x}{2}=0\)

\(\Leftrightarrow cos\frac{x}{2}\left(2sin\frac{x}{2}+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=0\\sin\frac{x}{2}=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

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

2.

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

\(\Leftrightarrow cosx=cos\left(x-\frac{\pi}{3}\right)\)

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

\(\Rightarrow x=\frac{\pi}{6}+k\pi\)

3.

\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=0\)

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

\(\Leftrightarrow...\)

4.

\(1+\frac{1}{2}sin6x=sin^2x+cos^2x+2sinx.cosx\)

\(\Leftrightarrow\frac{1}{2}sin6x=sin2x\)

\(\Leftrightarrow sin6x-2sin2x=0\)

\(\Leftrightarrow3sin2x-4sin^32x-2sin2x=0\)

\(\Leftrightarrow sin2x-4sin^32x=0\)

\(\Leftrightarrow sin2x\left(1-4sin^22x\right)=0\)

\(\Leftrightarrow sin2x\left(2cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)