(sinx)^2=2-2cosx

Question

Find all solutions of the equation in the interval [0,2pi).

 

(sinx)^2=2-2cosx

 

Write your answer in radians in terms of pi

Answer 1

Use the Pythagorean theorem identity as Kenneth showed to get the equation only in terms of cosx

 

1 - cos²x = 2 - 2cosx

 

 

Next, move all terms to the right side of the equation so that we have a positive coefficient of cos²x.  This will make it easier to factor using FOIL in the later steps.

 

0 = cos²x - 2cosx + 1

 

 

Now we factor using FOIL.

 

0 = (cosx - 1)(cosx - 1)

 

 

Since the factors are the same, we only need to set one of them equal to zero.

 

cosx - 1 = 0

 

cosx = 1

 

 

Now ask yourself this question.  On a unit circle, which angles have an x-coordinate of 1?  The angle will be your x value in pi radians.

 

 

Answer 2

Use the Pythagorean identity to substitute l - cos²x for sin²x.

You then have a quadratic equation; you could substitute u = cos x and solve for u, and then find out what x values that translated into.