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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?

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  1. anonymous
    • 5 years ago
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    ready?

  2. anonymous
    • 5 years ago
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    first we solve for sin(x) and then we have to solve for x.

  3. anonymous
    • 5 years ago
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    so let us call \[sin(x)=z\] so as not to confuse the algebra with the trig

  4. anonymous
    • 5 years ago
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    \[z^2=z\] \[z^2-z=0\] \[z(z-1)=0\] \[z+0\] or \[z=1\]

  5. anonymous
    • 5 years ago
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    typo should be \[z=0\] or \[z=1\]

  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    from the cheat sheet although if you are taking trig you should memorize these

  8. anonymous
    • 5 years ago
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    how did you get from z(z-1)= 0 to z=0

  9. anonymous
    • 5 years ago
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    nevermind i got it, and so coult another answer be pi then?

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