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    Alright, the question is to solve this equation.
    (3cos(2x))/(cos(2x)+1) = 2 ( tan(x) + 1 )

    I've tried for ages and i'm not really obtaining any solutions that when I put back in the equation really satisfy it. Can anyone help?

    Any help, is much appreciated. Tyy
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    (Original post by akash3141)
    Alright, the question is to solve this equation.
    (3cos(2x))/(cos(2x)+1) = 2 ( tan(x) + 1 )

    I've tried for ages and i'm not really obtaining any solutions that when I put back in the equation really satisfy it. Can anyone help?

    Any help, is much appreciated. Tyy
    Bear in mind that 2\cos^2(x) - 1 \equiv \cos(2x). Also it'll help to see your previous attempts.

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    also tagging to move to the right section Sonechka Lemur14
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    Also, as you have a tan(x) term, the other identity:

     \cos(2x) = \cos^{2}(x) - \sin^{2}(x),

    may be useful here as well as well as the identity provided by _gcx above.
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    (Original post by simon0)
    Also, as you have a tan(x) term, the other identity:

     \cos(2x) = \cos^{2}(x) - \sin^{2}(x),

    may be useful here as well as well as the identity provided by _gcx above.
    I got cos(x)[2cos(x)-4sin(x)] = 3
    idk if it's right or not but either way, I have no clue what to do after this, lol?
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    (Original post by akash3141)
    I got cos(x)[2cos(x)-4sin(x)] = 3
    idk if it's right or not but either way, I have no clue what to do after this, lol?
    - The equation is equaivalent to the question but not what I had in mind.

    - Starting again, for the left-hand equation try:
    1) My substitution for cos(2x) in the numerator,
    2) _gcx's substitution for cos(2x) in the denominator.
 
 
 
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