math help!!

Question

general solution, in degress, to the equation 2cosxcos2x-2sinxsin2x =-1

Answer 1

2cosxcos2x-2sinxsin2x =-1

or cosxcos2x-sinxsin2x =-1/2

 

Use the identity cos(A + B) = cosAcosB - sinA sinB

A=x, B=2x

Then, cosxcos2x-sinxsin2x = cos(x + 2x) = cos 3x = -1/2

Therefore, 3x = cos^-1 (-1/2) + 360 n

or x = [cos^-1 (-1/2) + 360 n]/3 = [cos^-1 (-1/2)]/3 + 120 n

The 360 n is necessary since the function repeats after every 360^o (n is an integer).  

Cos^-1(-1/2) = 120^o or 240^o.

Hence,

 x   = 40 + 120n    or x = 80 + 120n      degrees                     

Answer 2

You will apply the double-angle identities for sine and cosine to write the equation in another way.

 

2cosx(2cos²x - 1) - 2sinx(2sinxcosx) = -1

 

4cos³x - 2cosx - 4sin²xcosx = -1

 

 

Next, we can get everything in terms of cosine using the Pythagorean theorem identities, since the equation contains mostly cosine terms.

 

4cos³x - 2cosx - 4(1 - cos²x)cosx = -1

 

4cos³x - 2cosx - 4cosx + 4cos³x + 1 = 0

 

 

Now we combine like terms.

 

8cos³x - 6cosx + 1 = 0

 

 

 

Let           q=cosx

 

8q³ - 6q + 1 = 0

 

 

Solve for q using the graphing calculator.  The rational-root theorem will not help you here.  You will get three solutions of q.  Once you have your q values, substitute them back into

 

q=cosx

 

 

to find your three x values.