Answer
$\frac{\pi}{6}, \frac{5\pi}{6}, 3.3430, 6.0818$
Work Step by Step
Step 1. Use $cos^2(x)=1-sin^2x$; we have $10-10sin^2x+3sin(x)-9=0$ or $10sin^2x-3sin(x)-1=(2sin(x)-1)(5sin(x)+1)=0$, which gives $sin(x)=\frac{1}{2}$ and $sin(x)=-\frac{1}{5}$
Step 2. For $sin(x)=\frac{1}{2}$, the solutions in $[0,2\pi)$ are $x=\frac{\pi}{6}, \frac{5\pi}{6} $
Step 3. For $sin(x)=-\frac{1}{5}$, we can find the reference angle as $x_0=sin^{0.2}\approx0.2014$; thus the solutions in $[0,2\pi)$ are $x=\pi+x_0\approx3.3430$ and $x=2\pi-x_0\approx6.0818$
Step 4. The solutions to the original equation are $\frac{\pi}{6}, \frac{5\pi}{6}, 3.3430, 6.0818$
