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anonymous

  • one year ago

Help solving cos^2 (2x) -sin^2 (2x)=0 and sin(2x)=sin x ?

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  1. arindameducationusc
    • one year ago
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    Do you have options? i guess I found the first answer... but need to check if it matches @Kiritina

  2. anonymous
    • one year ago
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    cos^2 (2x) -sin^2 (2x) = cos(4x)

  3. anonymous
    • one year ago
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    No answers available, sorry. I'm not sure where to start with either equation.

  4. arindameducationusc
    • one year ago
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    can you manage now @Kiritina or should i do the solution?

  5. anonymous
    • one year ago
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    I'd love it if you helped with the solution!

  6. Jhannybean
    • one year ago
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    lets see... \[\cos^2(2x) -\sin^2(2x) = 0\]\[\color{red}{\sin(2x) = \sin(x)}\] is the red portion what you're given? Do you mind posting the actual problem by a picture?

  7. anonymous
    • one year ago
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    https://gyazo.com/976ec9f78a10f5487c8421aaa06e36a0

  8. arindameducationusc
    • one year ago
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    O yes @Jhannybean is right I just solved the black portion I didn't think about the red!

  9. arindameducationusc
    • one year ago
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    ok both are different problems hehe :D

  10. Jhannybean
    • one year ago
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    So you're looking for a general solution... x_x I'm terrible at finding those.

  11. arindameducationusc
    • one year ago
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    okay 1st question we have cos^2x-sin^2x=0 =>cos4x=0 \[=>4x=\cos^{-1} (0)\] \[=\cos^{-1} (\cos (\frac{ \pi }{ 2}))\] =>4x=pi/2 >x=pi/8

  12. arindameducationusc
    • one year ago
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    2nd question, sin2x=2sinxcosx so, sin2x=sinx =>2sinxcox/sinx=1 =>2cosx=1 =>cosx=1/2 x=cos inverse(1/2) x=cos inverse (cos pi/3) x=pi/3

  13. arindameducationusc
    • one year ago
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    @Kiritina if you liked my response, dont forget to medal and fan. Thank you

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