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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?
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first we solve for sin(x) and then we have to solve for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so let us call \[sin(x)=z\] so as not to confuse the algebra with the trig

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[z^2=z\] \[z^2z=0\] \[z(z1)=0\] \[z+0\] or \[z=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0typo should be \[z=0\] or \[z=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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 \[sin(\frac{\pi}{2})=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from the cheat sheet although if you are taking trig you should memorize these

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get from z(z1)= 0 to z=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nevermind i got it, and so coult another answer be pi then?
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