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anonymous

  • 5 years ago

Find all solutions on the interval [0,2pi] 2 cosx sinx - cos x =0 Please show+explain work?

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  1. anonymous
    • 5 years ago
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    did you get this yet? if not i will solve.

  2. watchmath
    • 5 years ago
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    go ahead :)

  3. anonymous
    • 5 years ago
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    first i would write \[2cos(x)sin(x)=cos(x)\] then square to get \[4cos^2(x)sin^2(x)=cos^2(x)\] \[4cos^2(x)[1-cos^2(x)]=cos^2(x)\] \[4cos^2(x)-4cos^4(x)-cos^2(x)=3cos^2(x)-4cos^4(x)=0\]

  4. anonymous
    • 5 years ago
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    how am i doing so far?

  5. watchmath
    • 5 years ago
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    If i do this I will move the cos x to the left and factor out the cos x

  6. anonymous
    • 5 years ago
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    \[4cos^4(x) - 3cos^2(x)=0\] \[cos^2(x)(4cos^2(x)-3)=0\] \[cos^2(x)=0\] or \[cos^2(x)=\frac{3}{4}\]

  7. anonymous
    • 5 years ago
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    \[cos(x) = 0\] \[x=\frac{\pi}{2}\]or \[x=-\frac{\pi}{2}\]

  8. anonymous
    • 5 years ago
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    proof 2 now that i am on a roll. \[f'(x)=3ax^2-4ax-8a\] the zeros are \[\frac{4a\pm\sqrt{16a^2+96a}}{6a}=\frac{4a\pm4\sqrt{a^2+6a}}{6a}=\] \[\frac{2\pm 2\sqrt{a(a+6)}}{3a}\] and in order for this to be -1 \[a(a+6)\] must be a perfect square, a miracle if ever there

  9. anonymous
    • 5 years ago
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    ignore that last gibberish it is wrong and not from here anyway.

  10. watchmath
    • 5 years ago
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    Alternative solution \(\cos x(2\sin x -1)=0\) \(\cos x=0\Rightarrow x=\pi/2,3\pi/2\) \(\sin x=1/2\Rightarrow x=\pi/6,5\pi/6\).

  11. anonymous
    • 5 years ago
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    btw we knew \[\pm\frac{\pi}{2}\]worked to begin with by inspection.

  12. anonymous
    • 5 years ago
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    right, much swifter and i retire now. yikes.

  13. anonymous
    • 5 years ago
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    Thanks!

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