Question

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

Answer 1

ready?

Answer 2

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

Answer 3

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

Answer 4

\[z^2=z\]
\[z^2-z=0\]
\[z(z-1)=0\]

\[z+0\] or
\[z=1\]

Answer 5

typo should be
\[z=0\]
or
\[z=1\]

Answer 6

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 \[sin(\frac{\pi}{2})=1\]

Answer 7

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

Answer 8

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

Answer 9

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