Question

Find all solutions to the equation. Show steps! Answer correctly for Brainliest!
sin^(2)x + sin x = 0

Answer 1

Answer:

$\displaystyle x=\left\{n\pi, \frac{3\pi}{2}+2n\pi \right\}, n\in\mathbb{Z}$

Step-by-step explanation:

We want to solve the equation:

$\sin^2x+\sin x=0$

Factor:

$\sin x(\sin x+1)=0$

Zero Product Property:

$\sin x=0\text{ or } \sin x+1=0$

Recall that sine equals 0 at 0 radians. This will repeat every π radians. So:

$x=n\pi, n\in\mathbb{Z}$

(Where n is an integer).

In the second case, we have:

$\sin x=-1$

This occurs at 3π/2 and will repeat every cycle or 2π. Hence:

$\displaystyle x=\frac{3\pi}{2}+2n\pi, n\in\mathbb{Z}$

Our solution is:

$\displaystyle x=\left\{n\pi, \frac{3\pi}{2}+2n\pi \right\}, n\in\mathbb{Z}$

Notes:

In the interval [0, 2π), the solutions are:

$\displaystyle x=\left\{0,\pi, \frac{3\pi}{2}\right\}$