Question

Find all solutions in the interval [0, 2π).
2 sin^2(x) = sin(x)

Answer 1

I will award a medal

Answer 2

set it to zero so

2sin^2(x) - sin(x) = 0

factorise

sin(x)[2sin(x) -1] = 0

you should be able to solve it from here

Answer 3

Well, let's see.
You want to find all the solutions for:
$$
2sin^2(x) = sin(x)
$$We could simplify that by dividing by sin(x).
The problem is that the division is defined only when \(sin(x) \ne 0\).
So let's first check what happens when \(sin(x) = 0\):
$$
2(0)^2 = 0 \\
0 = 0
$$
Means, when \(sin(x) = 0\) we have a solution. That means that:
$$
x = \pi\cdot K
$$ are some of the solutions of the equation.
Now, in case \(sin(x) \ne 0\) we can divide:
$$
2sin^2(x) = sin(x) \\
\frac{2sin^{\cancel2}(x)}{\cancel{sin(x)}} = \frac{\cancel{sin(x)}}{\cancel{sin(x)}} \\
2sin(x) = 1 \\
sin(x) = \frac{1}{2}
$$Means, when \(sin(x) = \frac{1}{2}\) we also have solutions for the equation.
$$
x = \frac{\pi}{6} + 2\pi \cdot K \\
x = \frac{5\pi}{6} + 2\pi \cdot K
$$
Limiting the solutions to the given interval:
$$
0 \le x \lt 2\pi \\
x = 0,\; x = \pi \;\;\;\;\;\;\;\;\;\text{for } sin(x) = 0 \\
x = \frac{\pi}{6},\; x = \frac{5\pi}{6}\;\;\;\;\text{for } sin(x) \ne 0 \\
$$
And we can also verify:
http://www.wolframalpha.com/input/?i=2sin%5E2%28x%29+%3D+sin%28x%29%2C+0%3C%3Dx%3C2pi