We'll recognize a linear equation and we'll solve it using the substitution technique.

We'll write sin a and cos a with respect to tangent of the half-angle.

sin a = 2 tan(a/2)/[1+(tan a/2)^2]

cos a = [1-(tan a/2)^2]/[1+(tan a/2)^2]

We'll substitute tan (x/2) = t

sin a = 2 t/(1+t^2)

cos a = (1-t^2)/(1+t^2)

We'll re-write the equation in t:

2*2 t/(1+t^2) + (1-t^2)/(1+t^2) = 2

We'll multiply by (1+t^2), the right side term 2:

4 t + (1-t^2) = 2(1+t^2)

We'll remove the brackets:

4t + 1 - t^2 = 2 + 2t^2

We'll move all terms to one side:

-3t^2 + 4t - 1 = 0

We'll multiply by -1:

3t^2 - 4t + 1 = 0

We'll apply the quadratic formula:

t1 = [4+sqrt(16-12)]/6

t1 = (4+2)/6

t1 = 1

t2 = (4-2)/6

t2 = 1/3

Now, we'll determine the angle a:

tan (a/2) = 1

a/2 = arctan 1 + k*pi

a = 2arctan 1 + 2k*pi

a = 2*pi/4 + 2k*pi

a = pi/2 + 2k*pi

tan (a/2) = 1/3

a = 2arctan (1/3) + 2k*pi

**All the values of the angle a, that make the identity true, are: ****{pi/2 + 2k*pi} U {2arctan (1/3) + 2k*pi}.**