7.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)
\(\Leftrightarrow\dfrac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=sinx+cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\Rightarrow x=-\dfrac{\pi}{4}+k\pi\\\dfrac{cosx-sinx}{sinx.cosx}=1\left(1\right)\end{matrix}\right.\)
Xét (1)
\(cosx-sinx=sinx.cosx\)
Đặt \(cosx-sinx=t\) với \(\left|t\right|\le\sqrt{2}\)
\(\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)
\(\Rightarrow t=\dfrac{1-t^2}{2}\)
\(\Leftrightarrow t^2+2t-1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=-1+\sqrt{2}\)
\(\Rightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{2-\sqrt{2}}{2}\)
\(\Rightarrow x=-\dfrac{\pi}{4}\pm arccos\left(\dfrac{2-\sqrt{2}}{2}\right)+k2\pi\)
8.
\(1+cos2x+cosx+cos3x=0\)
\(\Leftrightarrow2cos^2x+2cos2x.cosx=0\)
\(\Leftrightarrow cosx\left(cosx+cos2x\right)=0\)
\(\Leftrightarrow cosx\left(2cos^2x+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
