Trigonometry problem help

  • Thread starter ritwik06
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  • #1
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Homework Statement



Find the solution of
2(cos x + cos 2x)+sin 2x(1+2cos x)=2 sin x in the interval [-pi, pi]


The Attempt at a Solution


Simplifying:
2 (cos x- sin x)+ sin 2x +cos 2x +2sin 2x cos x=0
2 (cos x- sin x)+ sin 2x +cos 2x +sin 3x+sin x=0
cos 2x+2cos x+sin 2x-sin x+sin 3x=0

But now its difficult to convert into factors to apply zero prouct rule.
 

Answers and Replies

  • #2
HallsofIvy
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I would recommend you get rid of "sin 2x" and "cos 2x" by using sin 2x= 2sin x cos x and cos 2x= cos2 x- sin2 x.
 
  • #3
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I would recommend you get rid of "sin 2x" and "cos 2x" by using sin 2x= 2sin x cos x and cos 2x= cos2 x- sin2 x.

Changing all the multiples of x, I get:
cos2x+2 cos x +2 sin x cos x -4 sin3x+2sin x -sin 2x=0

cos x(cos x+sin x +2)+sin x(2cos 2x+sin x)=0

Again I am stuck?????
 

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