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    Solve 5 sin 4x = 2 sin 2x in the interval 0<x<360

    5 (sin 2xcos2x + cos2xsin2x) -2sin2x = 0
    5 (2sin 2x cos2x) -2(sinx cosx) = 0
    10 sin2x cos 2x- 4 sinx cos x =0
    [10 (2 sinx cos x) (1-2sin2x)]- 4 sinx cos x = 0
    20 sinx cosx - 40 sin^3 x cos x - 4 sinx cos x= 0
    16 sinx cosx- 40 sin^3 x cos x = 0
    8sinx cosx (2-5 sin^2 x) = 0

    8 sinx cos x = 0

    tan x = 0

    tan^(-1) = 0, 180, 360

    2- sin^2 x= 0
    sin x = +- root (2/5)
    = 23.6, 156.4

    Is this correct, and is there a shorter way?
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    I would keep it as 10sin2xcos2x-2sin2x=0 so 2sin2x(5cos2x-1)=0
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    To check if your solutions are correct, substitute them back into the original equation (it's always worth doing this in an exam in case you've made some small arithmetic errors).

    I've been beaten to it but here's what I was going to post. The above should still be handy anyway:

    Spoiler:
    Show

    I wouldn't have bothered turning everything into sin(x) and cos(x), instead at the first step you have

     5(2\sin 2x \cos 2x) - 2 \sin 2x = 0

    So this is  2\sin 2x (5 \cos 2x - 1) = 0
    Hence  \sin 2x = 0 or  \cos 2x = \frac{1}{5} .
    The first you should be able to solve without a calculator. The second might require a calculator.
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    Of course, thanks!

    Also, I wasn't sure when I was doing this part (should've done it your way)
    but does 8 sinx cos x = 0 become tan x = 0 , is it right?
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    (Original post by calamityx)
    Of course, thanks!

    Also, I wasn't sure when I was doing this part (should've done it your way)
    but does 8 sinx cos x = 0 become tan x = 0 , is it right?
    You divided through by 8cos^2x? You need to be careful because that could equal zero.A counter example to what you have said is 90 degrees. 8sin90cos90=0 but tan90 is undefined.

    I would be inclined to rewrite that as 4sin2x=0sosin2x=0
 
 
 
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