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    Can anybody help me solve this equation:

    sin2x - tanx = 2cos2x
    for x in the interval 0<= x <= 180 degrees. (Greater than or equal to 0 and less than or equal to 180 degrees, if unclear).

    So far, I have:
    4cosx - 4cos^3 + 2sinxcos^2 -sinx -2cosx = 0

    Thanks in advance
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    (Original post by UserXi1.fr)
    Can anybody help me solve this equation:

    sin2x - tanx = 2cos2x
    for x in the interval 0<= x <= 180 degrees. (Greater than or equal to 0 and less than or equal to 180 degrees, if unclear).

    So far, I have:
    4cosx - 4cos^3 + 2sinxcos^2 -sinx -2cosx = 0

    Thanks in advance
    Try this

    2sincos^2 - sin = 2cos(2cos^2-1)

    Which is a step before where you were

    Factorise the LHS
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    These were my individual steps:
    1)sin2x - tanx = 2cos2x
    2)2sinxcosx - tanx = 2(1- 2sin^2x)
    3) 2sinxcosx -tanx -2 +4sin^2x = 0
    4) 4sin^2 + 2sinxcosx - tanx -2 = 0
    5) 4sin^2 + 2sinxcosx - sinx/cosx -2 = 0
    6) 4sin^2 cosx + 2sinxcos^2 - sinx - 2cosx = 0
    7) 4(1-cos^2)cosx +2sinxcosx -sinx -2cosx =0
    8) 4cosx -4cos^3 +2sinxcos^2 -sinx -2cosx= 0

    That's as far as I got, but with a variety of sinx and cosx, any obvious errors?
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    I got 2sin^2(x)-sin(x)-1=0

    is this correct?
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    tan x cos 2x = 2 cos 2x
    cos 2x (tan x − 2) = 0 M1
    cos 2x = 0 or tan x = 2 A1
    2x = 90, 270 or x = 63.4 B1
    x = 45°, 63.4° (1dp), 135°

    That is the official answer in the mark scheme
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    (Original post by UserXi1.fr)
    These were my individual steps:
    1)sin2x - tanx = 2cos2x
    2)2sinxcosx - tanx = 2(1- 2sin^2x)
    3) 2sinxcosx -tanx -2 +4sin^2x = 0
    4) 4sin^2 + 2sinxcosx - tanx -2 = 0
    5) 4sin^2 + 2sinxcosx - sinx/cosx -2 = 0
    6) 4sin^2 cosx + 2sinxcos^2 - sinx - 2cosx = 0
    7) 4(1-cos^2)cosx +2sinxcosx -sinx -2cosx =0
    8) 4cosx -4cos^3 +2sinxcos^2 -sinx -2cosx= 0

    That's as far as I got, but with a variety of sinx and cosx, any obvious errors?
    If you use 2cos^2 - 1 instead in line 2 you have what I had which factorised as the mark scheme shows
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    is this edxcel,if so which paper
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    http://www.freeexampapers.com/past_p...FC3%2FSolomon/

    Paper E, Q 2B
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    (Original post by TenOfThem)
    If you use 2cos^2 - 1 instead in line 2 you have what I had which factorised as the mark scheme shows
    I've used that... yet to no avail. Any complete solutions for the answer, mark scheme seems to skip a few steps.
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    (Original post by UserXi1.fr)
    I've used that... yet to no avail. Any complete solutions for the answer, mark scheme seems to skip a few steps.
    Do you have what I put in post 2

    Can you factorise the LHS
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    2)2)2sinxcosx - tanx = 2(1- 2sin^2x)

    Yeah so if I replaced (1-2sin^2 x) with (2cos^2 x -1), so

    2sinxcosx - sinx/cosx = 2( 2cos^2 x -1)
    2sinxcos^2 - sinx = 2(2cos^2 -1).

    Stuck here, tried it multiple times, end up expanding and getting cosx and sinx all over the place!
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    Got 2/3 solutions.

    1) 2sincos^2 -sinx= 2(2cos^2 -1)
    2) sinx(2cos^x-1) = 2(2cos^2-1)
    3) sinxcos2x -2cos2x = 0
    4) cos2x( sinx-2)=0
    5) cos2x=0 {sinx=2... no solutions}
    6) 2x= 90, 270.....x=45, 135.
    Missing a 63.4 degrees though.
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    (Original post by UserXi1.fr)
    Got 2/3 solutions.

    1) 2sincos^2 -sinx= 2(2cos^2 -1)
    2) sinx(2cos^x-1) = 2(2cos^2-1)
    3) sinxcos2x -2cos2x = 0
    4) cos2x( sinx-2)=0
    5) cos2x=0 {sinx=2... no solutions}
    6) 2x= 90, 270.....x=45, 135.
    Missing a 63.4 degrees though.
    You are missing a cos in line 2
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    (Original post by UserXi1.fr)
    Can anybody help me solve this equation:

    sin2x - tanx = 2cos2x
    for x in the interval 0<= x <= 180 degrees. (Greater than or equal to 0 and less than or equal to 180 degrees, if unclear).
    I think the easiest method is to form a cubic in tan x.

    \sin 2x -\tan x = 2 \cos 2x

    2 \sin x \cos x - \tan x = 2 \cos^2 x - 2\sin^2 x

    Divide each term by \cos^2 x

    2 \tan x- \tan x \sec^2 x = 2 - 2\tan^2 x

    Use the Pythagorean identity \sec^2 x = 1 + \tan^2 x and you should be home.
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    (Original post by Mr M)
    I think the easiest method is to form a cubic in tan x.

    \sin 2x -\tan x = 2 \cos 2x

    2 \sin x \cos x - \tan x = 2 \cos^2 x - 2\sin^2 x

    Divide each term by \cos^2 x

    2 \tan x- \tan x \sec^2 x = 2 - 2\tan^2 x

    Use the Pythagorean identity \sec^2 x = 1 + \tan^2 x and you should be home.

    Good, found the solutions using both methods.
    Thanks all!
 
 
 
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