Find the sum of all solutions of equation (cosx) ^ 2 - (SiNx) ^ 2 = 1 / 2 in the interval [- 2 π, 2 π]

Find the sum of all solutions of equation (cosx) ^ 2 - (SiNx) ^ 2 = 1 / 2 in the interval [- 2 π, 2 π]

(cosx)^2-(sinx)^2=1/2
cos2x=1/2
Because x ∈ [- 2 π, 2 π]
Then 2x ∈ [- 4 π, 4 π]
Because y = cos2x is an even function
All the points of y = cos2x = 1 / 2 are symmetrically distributed about the Y axis
The given interval [- 2 π, 2 π] is also symmetric about the y-axis
So the sum of all solutions is 0

The sum of all solutions of the equation SiNx + cosx = (√ 2) / 2 in the interval [0,4 Π] is ()?
7∏
Is there a way to see it directly from the picture?

sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2(sinxcos∏/4+cosxsina∏/4)=(√2)sin(x+∏/4)==(√2)/2
So sin (x + Π / 4) = 1 / 2, and because x is on [0,4 Π], so x can be 7 Π / 12,23 Π / 12,31 Π / 12,47 Π / 12, which adds up to 9 Π