1. arcsin(3x -pi) = 1/2
2. arcsin rad(2x) = arccos rad(x)
1. as mentioned earlier, note that ...
$\displaystyle -1 \le 3x-\pi \le 1$
therefore the value of x is restricted to ...
$\displaystyle \frac{\pi - 1}{3} \le x \le \frac{\pi + 1}{3}$
solving the equation ...
$\displaystyle 3x - \pi = \sin\left(\frac{1}{2}\right)$
$\displaystyle x = \frac{1}{3}\left[\sin\left(\frac{1}{2}\right) + \pi\right] \approx 1.207$
2. since arcsin(something) = arccos(something else) , the angle these two expressions equal can only be in quad I (why is that?)
let $\displaystyle \theta = \arcsin{\sqrt{2x}}$
$\displaystyle \cos{\theta} = \sqrt{1-2x}$
since the angles are equal ...
$\displaystyle \theta = \arccos{\sqrt{x}}$
$\displaystyle \cos{\theta} = \sqrt{x}$
so ...
$\displaystyle \sqrt{1-2x} = \sqrt{x}$
solve for $\displaystyle x$