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Mathematics 7 Online
OpenStudy (anonymous):

sin2x-sinx=0

OpenStudy (anonymous):

what should we do ? prove it or what ?

OpenStudy (jonnyvonny):

To find the zeroes, you may factor out a "sinx" from both go get:\[sinx(sinx-1)=0\] This tells us that the values that make y zero are when sinx=0, and sinx=1 (made sinx alone inside the parenthesis). From our unit circle, sinx=0 at 0 and 2pi. Sinx=1 at pi/2 and 3pi/2. If you need more elaboration, feel free to ask.

OpenStudy (anonymous):

i think sin2x = 2sinxcosx not 2sinx

OpenStudy (anonymous):

sin2x-sinx=0 2sinxcosx-sinx=0 sinx(2cosx-1)=0 u can find the value of x sinx=0 2cosx-1=0

OpenStudy (jonnyvonny):

Mmm, indeed, depends if she meant sin^2(x) or sin(2x)...I believe you're correct in the interpretation, however.

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