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anonymous

  • 5 years ago

Find all the values of x in the interval [0, 2pie] that satisfy the equation 2cos(x) + sin(2x) = 0.

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  1. anonymous
    • 5 years ago
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    Substitute sin 2x. Here's your cheat sheet http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

  2. anonymous
    • 5 years ago
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    Can you show me the full solution please?

  3. anonymous
    • 5 years ago
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    Try my suggestion, I would correct you if you do something wrong.

  4. anonymous
    • 5 years ago
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    sin(2x)=2sin(x)cos(x) then you get 2cos(x)+2cos(x)sin(x)=0 2cos(x)(1+sin(x))=0 cos(x)=0 or 1+sin(x)=0 sin(x)=−12

  5. anonymous
    • 5 years ago
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    -1/2*

  6. anonymous
    • 5 years ago
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    The two different quantities in brackets are multiplied; or rather the three quantities are multiplied, so when evaluating each the other two go to zero. The 2 can actually be ignored. So cos(x)=0 1 + sin x=0 sin x= -1

  7. anonymous
    • 5 years ago
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    in that interval cosine is 0 at π2 and 3π2

  8. anonymous
    • 5 years ago
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    sin(x)=−1/2 and see that it is at 7π/6 and 11π/6

  9. anonymous
    • 5 years ago
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    \[x =\cos^{-1} 0\]\[x =\pi/2,(3\pi/2)\]\[x =\sin^{-1} -1\]\[x =(3\pi/2)\](Remember there is no -1/2. The 2 goes to 0.

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