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Hannah_Ahn

  • 4 years ago

solve: sin^2x=sinxcosx

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  1. ash2326
    • 4 years ago
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    sin^2x =sin x cosx sin^2 x-sinx cos x=0 sinx (sin x-cos x)= either sin x=0 or sin x = cos x sin x =0 so \[x= n \pi\] sin x = cos x \[x= 2n \pi + \pi/4\]

  2. Hannah_Ahn
    • 4 years ago
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    The answer is x=0, pi over 4, pi, 5pi over 4. and how would I get 5pi over 4? Thanks.

  3. ash2326
    • 4 years ago
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    yeah sorry,it'll also include \[n \pi+ \pi/4\]

  4. ash2326
    • 4 years ago
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    so you'll get, 0 , pi/4 and 5pi/4

  5. Hannah_Ahn
    • 4 years ago
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    I think I got it. Thanks.

  6. ash2326
    • 4 years ago
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    welcome

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