Drilon Posted January 11, 2007 Share Posted January 11, 2007 I started to solve this equation but it seems I'm doing something wrong. Here s the equation: cos^3 x-3cos^2 x+ cos x = 2cos(x/2 + pi/4)sin(3x/2 - pi/4) I got the first solution X1= (2n+1)pi/2 The second solution according to the book: X2=2pi*n Link to comment Share on other sites More sharing options...

Ragib Posted January 12, 2007 Share Posted January 12, 2007 I can't be bothered to, could you expand 2cos(x/2 + pi/4)sin(3x/2 - pi/4), take the terms to one side and make any simplifications you can. Then I'll help you find the other solutions. Link to comment Share on other sites More sharing options...

Drilon Posted January 13, 2007 Author Share Posted January 13, 2007 I can't be bothered to, could you expand 2cos(x/2 + pi/4)sin(3x/2 - pi/4), take the terms to one side and make any simplifications you can. Then I'll help you find the other solutions. Sorry I didn't do it earlier. Ill do it now. cos^3 x-3cos^2 x+ cos x = 2cos(x/2 + pi/4)sin(3x/2 - pi/4) cos^3 x-3cos^2 x+ cos x= sin2x - cos x cos^3 x-3cos^2 x+ 2cos x - 2sinxcosx=0 cos x(cos^2 - 3cos x +2 - 2sin x)=0 if a*b=0 then a=0 and b=0 cos x=0 => X1=(2n+1)pi/2 and the second equation is: cos^2x - 3cos x - 2sin x = -2 the equation matches the result X2=2n*pi but I just can't figure out how to solve it because everything I do just messes up and makes it more complicated. P.S. Thanks in advance. Link to comment Share on other sites More sharing options...

Ragib Posted January 14, 2007 Share Posted January 14, 2007 Well since there is both sin and cos in one equation, all i can think of is t substitution. Let t=tan(x/2), since you know the the expansion for tan(x+y), let x=x/2 and y=x/2, that way tan x= 2t/(1-t^2). Right a right triangle, set 2t the opposite side, 1-t^2 the adjacent, use pythagoras for the remaining sides. Then the opp/hypotenuse ratio is sin x, and do the same for cos. Make the substuitions for cos and sin into the last equation, solve for t and you can get a quadratic equation, which is easy to solve. Good luck Link to comment Share on other sites More sharing options...

Drilon Posted January 14, 2007 Author Share Posted January 14, 2007 Well thanks again. Link to comment Share on other sites More sharing options...

Ragib Posted January 15, 2007 Share Posted January 15, 2007 Did you get it? Link to comment Share on other sites More sharing options...

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