a, \(4sin^3x+4sin^2x=3+3sinx\)
\(\Leftrightarrow\left(sinx+1\right)\left(4sin^2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+1=0\\4sin^2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+1=0\\1+2cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{2}+k2\pi\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
b, \(sin3x+1=2sin^2x\)
\(\Leftrightarrow3sinx-4sin^3x+1=2sin^2x\)
\(\Leftrightarrow4sin^3x+2sin^2x-3sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\dfrac{1-\sqrt{5}}{4}\\sinx=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)
TH1: \(sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)
TH2: \(x=\dfrac{1-\sqrt{5}}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\dfrac{1-\sqrt{5}}{4}\right)+k2\pi\\x=\pi-arcsin\left(\dfrac{1-\sqrt{5}}{4}\right)+k2\pi\end{matrix}\right.\)
TH3: \(x=\dfrac{1+\sqrt{5}}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\dfrac{1+\sqrt{5}}{4}\right)+k2\pi\\x=\pi-arcsin\left(\dfrac{1+\sqrt{5}}{4}\right)+k2\pi\end{matrix}\right.\)
c, \(1+sin3x-sinx=cos2x\)
\(\Leftrightarrow1+2cos2x.sinx=1-2sin^2x\)
\(\Leftrightarrow2\left(cos2x+sinx\right).sinx=0\)
\(\Leftrightarrow\left(2sin^2x-sinx-1\right).sinx=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sinx+1\right)sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=0\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
TH1: \(sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)
TH2: \(sinx=0\Leftrightarrow x=k\pi\)
TH3: \(sinx=-\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\pi+\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
