Question

Find all solutions to the equation.sin^2x + sin x = 0Thanks in advance!

Answer 1

Given:

`sin^2x+sinx=0`

To solve the function, factor the function using sin(x):

`sin(x)*[sin(x)+1\rbrack=0`

That means, in order to the produtct be zero, or sin(x) or [sin(x)+ 1] = 0

So, let's solve each situation.

`\begin{gathered} sin(x)=0 \\ By\text{ taking the inverse:} \\ x=sin^{\left\{-1\right\}}\left(0\right) \\ x=\pi n,\text{ n is a integer} \end{gathered}`

Or:

`\begin{gathered} sin(x)+1=0 \\ \text{ Subtracting 1 from both sides:} \\ sin(x)+1-1=0-1 \\ sin(x)=-1 \\ \text{ Taking the inverse:} \\ x=sin^(-1)(-1) \\ x=(3)/(2)\pi+2\pi n \end{gathered}`

Answer:

x = πn, 3/2 π + 2πn, for n integer n.