Giải thích các bước giải:

Vậy 

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

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

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

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

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

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

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

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

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

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

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

Đk: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m2\pi\\x\ne\dfrac{\pi}{4}+n\pi\end{matrix}\right.\left(m,n\in Z\right)\)

PT \(\Leftrightarrow1=2\sqrt{2}sinx.cosx\left(sinx-cosx\right)+2cos^2x\)

\(\Leftrightarrow\sqrt{2}.2sinx.cosx\left(sinx-cosx\right)+\left(2cos^2x-1\right)=0\)

\(\Leftrightarrow\sqrt{2}sin2x\left(sinx-cosx\right)+\left(cosx-sinx\right)\left(cosx+sinx\right)=0\)

\(\Leftrightarrow\sqrt{2}sin2x=sinx+cosx\)

\(\Leftrightarrow\sqrt{2}sin2x=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)

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

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

Cảm mơn nhiều nha :3

Đặt \(cosx-sinx=t\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)

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

Pt trở thành:

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

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

\(\Leftrightarrow\left(t-2\right)\left(t+1\right)^2=0\Rightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=-1\end{matrix}\right.\)

\(\Rightarrow cosx-sinx=-1\)

\(\Leftrightarrow\sqrt[]{2}cos\left(x+\dfrac{\pi}{4}\right)=-1\)

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow...\)

Dạ em cảm ơn ạ!! ^^

Lời giải:

$m^2=(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x=1+2\sin x\cos x$

$\Rightarrow \sin x\cos x=\frac{m^2-1}{2}$

Ta có:

$|\sin ^3x-\cos ^3x|=|\sin x-\cos x||\sin ^2x+\sin x\cos x+\cos ^2x|$

$=\sqrt{(\sin x-\cos x)^2}|1+\sin x\cos x|$

$=\sqrt{1-2\sin x\cos x}.|1+\sin x\cos x|$

$=\sqrt{1-(m^2-1)}.|1+\frac{m^2-1}{2}|$

$=\sqrt{2-m^2}.\frac{m^2+1}{2}$

\(sinx+cosx=m\\ \Rightarrow sin^2x+cos^2x+2sinx.cosx=m^2\\ \Rightarrow sinx.cosx=\dfrac{1-m^2}{2}\)

Mặt khác: 

\(sinx-cosx=\left(sinx+cosx\right)-2cosx=m-2cosx\)

Có:

\(\left|sin^3x-cos^3x\right|=\left|\left(sinx-cosx\right)\left(sin^2x+sinx.cosx+cos^2x\right)\right|\\ =\left|\left(m-2cosx\right)\left(1+\dfrac{1-m^2}{2}\right)\right|\\ =\left|\left(m-2cosx\right)\left(\dfrac{3-m^2}{2}\right)\right|\)

\(y=\dfrac{sinx+1}{sinx}\)

ĐKXĐ: \(sinx\ne0\Rightarrow x\ne k\pi\)

\(y=\dfrac{sin2x+cosx}{tanx-sinx}\)

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\tanx-sinx\ne0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}cosx\ne0\\sinx\left(\dfrac{1}{cosx}-1\right)\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}cosx\ne0\\sinx\ne0\\cosx\ne1\end{matrix}\right.\)

\(\Rightarrow sin2x\ne0\)

\(\Rightarrow x\ne\dfrac{k\pi}{2}\)