## anonymous 5 years ago in the interval 0 less than or equal to x greater than 2pi the solutions of the equation sin^2x=sin x are?

1. anonymous

2. anonymous

first we solve for sin(x) and then we have to solve for x.

3. anonymous

so let us call $sin(x)=z$ so as not to confuse the algebra with the trig

4. anonymous

$z^2=z$ $z^2-z=0$ $z(z-1)=0$ $z+0$ or $z=1$

5. anonymous

typo should be $z=0$ or $z=1$

6. anonymous

now replace z by $sin(x)$ and solve $sin(x)=0$ to get $x=0$ because $sin(0)=0$ or $sin(x)=1$ to get $x=\frac{\pi}{2}$ because $sin(\frac{\pi}{2})=1$

7. anonymous

from the cheat sheet although if you are taking trig you should memorize these

8. anonymous

how did you get from z(z-1)= 0 to z=0

9. anonymous

nevermind i got it, and so coult another answer be pi then?