$2 \cos(5x+10)=1$

$\cos(5x+10)=\frac{1}{2}$

$5x+10=\arccos\left(\frac{1}{2}\right)$

for $0 \leq x < 360$ there are two values of $\arccos\left(\frac{1}{2}\right)$, $x=60$ and $x=300$ degrees

$5x+10=60$

$5x=50$

$x=10$

$5x+10=300$

$5x=290$

$x = 58$

The second problem is similar, just be careful taking the arcsin.