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 C.

TH1: Nếu cosx =0 có sin²x = 1 không thỏa mãn phương trình.

TH2:  chia cả hai vế của phương trình cho cos²x ta được:

Đáp án D

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:

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

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

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

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

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

\(\Rightarrow x=...\)

cos2x = 1- sin^x 
sin2x= 2sinxcosx 

Nhóm lại bình thường và giải thôi

ĐK: \(x\ne\dfrac{\pi}{6}+k2\pi;x\ne\dfrac{5\pi}{6}+k2\pi\)

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

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

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

Đối chiếu điều kiện ta được \(x=-\dfrac{5\pi}{6}+k2\pi\).