Answer:
x = {kπ, arccos(1/5) +2kπ, 2kπ -arccos(1/5)}
Explanation:
The double-angle trig identity for sine is useful:
5(2sin(θ)cos(θ)) -2sin(θ) = 0
2sin(θ)(5cos(θ) -1) = 0
This has solutions that make the factors zero:
θ = arcsin(0) = kπ
and ...
cos(θ) = 1/5
θ = arccos(1/5) +2kπ . . . . or . . . . 2kπ -arccos(1/5)
_____
Some numerical values are shown on the graph attached. values for multiples of pi are ...
{..., -12.566, -9.425, -6.283, -3.142, 0, 3.142, 6.283, 9.425, 12.566, ...}