1.
\(\Leftrightarrow sin5x=1-2cos^2x\)
\(\Leftrightarrow sin5x=-cos2x\)
\(\Leftrightarrow sin5x=sin\left(2x-\dfrac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=2x-\dfrac{\pi}{2}+k2\pi\\5x=\dfrac{3\pi}{2}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{3\pi}{14}+\dfrac{k2\pi}{7}\end{matrix}\right.\)
2.
\(\Leftrightarrow3sin^2x-7sinx.cosx+4cos^2x=0\)
Nhận thấy \(cosx=0\) ko phải nghiệm
Với \(cosx\ne0\) chia 2 vế cho \(cos^2x\)
\(\Rightarrow3tan^2x-7tanx+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{4}{3}\right)+k\pi\end{matrix}\right.\)