Solve sin2x+cos2x=sqrt(2) on the interval [0,2p)

Question

I've been stumped on this problem... Step-by-step please :)

Answer 1

Let        q=2x

 

sinq + cosq = √2

 

 

Get the equation in terms of either sine of cosine.

 

sinq + √[1 - sin²q] = √2

 

 

Square both sides of the equation.

 

sin²q + 2sinq√(1 - sin²q) + (1 - sin²q) = 2

 

 

Combine like terms where necessary.

 

2sinq√(1 - sin²q) + 1 = 2

 

2sinq√(1 - sin²q) = 1

 

 

Square both sides of the equation again to get rid of the radical sign.

 

4sin²q(1 - sin²q) = 1

 

4sin²q - 4sin⁴q - 1 = 0

 

 

Move all terms to the right side of the equation to make factoring easier.

 

0 = 4sin⁴q - 4sin²q + 1

 

 

Now we can try to factor using FOIL.

 

0 = (2sin²q - 1)(2sin²q - 1)

 

 

Next, set the factor equal to zero an solve for q.

 

 

2sin²q - 1 = 0

 

 

 

Then once you solve for q, substitute that value of q into the equation     q = 2x      to solve for x.