Question

2sin^2x-sinx=0

Answer 1

$$2 \sin^ {2} x - \sin x = 0$$ $$\sin x (2 \sin x - 1) = 0$$

The product is zero if one of the factors is equal to zero:

$$\left[ \begin{array}{l} \sin x = 0 \\ \sin x = 0.5 \end{array} \right.$$

We know the following formula:

$$\sin x = a, |a| \leq 1$$ $$x = (-1)^n \operatorname{asin} a + \pi n; n \in \mathbb{Z}$$ $$\left[ \begin{array}{c} x = \pi n, \quad n \in \mathbb{Z} \\ x = (-1)^n \operatorname{asin}(0.5) + \pi n = (-1)^n * \frac{\pi}{6} + \pi n; \end{array} \right.$$

We can show solutions on the unit circle