a. ĐKXĐ: ...
Ta có: \(\left\{{}\begin{matrix}VT=\left(tanx-cotx\right)^2+2\ge2\\VP=1+cos^2\left(3x+\frac{\pi}{4}\right)\le2\end{matrix}\right.\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}tanx-cotx=0\\cos^2\left(3x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}cos2x=0\\sin\left(3x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
b.
\(\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi\)
\(\Leftrightarrow sinx-1=3k\)
\(\Leftrightarrow sinx=3k+1\)
Do \(-1\le sinx\le1\)
\(\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0\)
\(\Rightarrow k=0\)
\(\Rightarrow sinx=1\)
\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
c.
ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{4}\left(cosx-1\right)=-\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow cosx-1=4k-1\)
\(\Leftrightarrow cosx=4k\)
Mà \(-1\le cosx\le1\Rightarrow-1\le4k\le1\)
\(\Rightarrow-\frac{1}{4}\le k\le\frac{1}{4}\Rightarrow k=0\)
\(\Rightarrow cosx=0\)
\(\Rightarrow x=\frac{\pi}{2}+k\pi\)
