Trig equation

[Drilon]

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

[Ragib]

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.

[Drilon]

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.

[Ragib]

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

[Drilon]

Well thanks again.

[Ragib]

Did you get it?