Question

What are all the exact solutions of 2 sin^2 x- sin x = 0

Answer 1

2sin^2 x - sinx = 0
add sinx to both sides

2sin^2 x = sinx
divide bot sides by sinx

2sinx = 1
divide both sides by 2

sinx = 1/2
inverse trig function

x = π/6
** but the sine function is harmonic and hits 1/2 every 2π from π/6

So the answer is..
x = π/6 + 2nπ

where n is an integer

Answer 2

Factor:

2sin^2x-sinx = 0

sinx(2sinx - 1) = 0

Therefore the solutions are when:

sin x = 0
And
sinx = 1/2

So sinx = 0
is true when x = 0 and pi  and all the angles coterminal with these points.  Thus, the answer is x = pi*n, (where n is some integer)

sinx = 1/2
is true when x = pi/6 and 5pi/6 and the angles coterminal with these points.
Thus, the answer is x = pi/6 + 2pi*n (where n is some integer)
and x = 5pi/6 + 2pi*n (where n is some integer)