ĐKXĐ; ...
\(\Leftrightarrow\frac{3}{5}sinx+\frac{4}{5}cosx-1=\frac{1}{5}\left(4tanx-3\right)^2\)
\(\Leftrightarrow sin\left(x+a\right)-1=\frac{1}{5}\left(4tanx-3\right)^2\)
(Trong đó \(a\in\left(0;\pi\right)\) sao cho \(cosa=\frac{3}{5}\))
Do \(\left\{{}\begin{matrix}sin\left(x+a\right)-1\le0\\\left(4tanx-3\right)^2\ge0\end{matrix}\right.\) \(\forall a;x\) nên đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sin\left(x+a\right)=1\\4tanx-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3sinx+4cosx=5\\4sinx-3cosx=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sinx=\frac{3}{5}\\cosx=\frac{4}{5}\end{matrix}\right.\) \(\Rightarrow x=arcsin\left(\frac{3}{5}\right)+k2\pi\)
