SOLUTION: 3cos^2 2x - 2sin^2 2x = 2 - 4sin2x find all the angles between 0degrees to 180degrees inclusive which satisfy equation --x-x-x-x-x-x-x-x-x-- express 1/ (x-3)(2x +1) as

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: 3cos^2 2x - 2sin^2 2x = 2 - 4sin2x find all the angles between 0degrees to 180degrees inclusive which satisfy equation --x-x-x-x-x-x-x-x-x-- express 1/ (x-3)(2x +1) as       Log On

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Question 302828: 3cos^2 2x - 2sin^2 2x = 2 - 4sin2x
find all the angles between 0degrees to 180degrees inclusive which satisfy equation

--x-x-x-x-x-x-x-x-x--
express 1/ (x-3)(2x +1) as partial fractions
hence evaluate (cant really draw it out but.... evaluate definite integral 7 on top, 4 below. 1/(x-3)(2x +1)
Answer by Fombitz(13828) About Me  (Show Source):
You can put this solution on YOUR website!
Let S=sin%282x%29 and C=cos%282x%29 to reduce the clutter,
3C%5E2-2S%5E2+=+2+-+4S
3C%5E2-2S%5E2%2B4S-2+=0
Remember that
C%5E2%2BS%5E2=1
C%5E2=1-S%5E2
3C%5E2=3-3S%5E2
Substitute,
%283-3S%5E2%29-2S%5E2%2B4S-2=0
-5S%5E2%2B4S%2B1=0
5S%5E2-4S-1=0
You can factor,
%285S%2B1%29%28S-1%29=0
Two solutions:
5S%2B1=0
S=-1%2F5
sin%282x%29=-1%2F5
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.
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S-1=0
S=1
sin%282x%29=1
I leave those to you to complete.
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.
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Partial fractions.
1%2F+%28%28x-3%29%282x+%2B1%29%29+=A%2F%28x-3%29%2BB%2F%282x%2B1%29
Multiply by %28x-3%29%282x%2B1%29
1=A%282x%2B1%29%2BB%28x-3%29
%282Ax%2BBx%29%2B%28A-3B%29=1
Leads to two equations,
1.2A%2BB=0
2.A-3B=1
Multiply eq.1 by 3 and add the two together to solve for A.
6A%2B3B%2BA-3B=1
7A=1
A=1%2F7
Then use either equation to solve for B.
2A%2BB=0
B=-2A=-2%2F7

If you integrate you'll get
INT=%281%2F7%29ln%28x-3%29-%282%2F7%29ln%282x%2B1%29 evaluated between x=4 and x=7.
I leave those to you to complete.