Solve the following trigonometric equation in the interval 0 < x < 2pi:

sin^2(x) - cos^2(x) = - cos(x)

Notice that sin^2(x)  = 1 - cos^2(x)    ...so we have

1- cos^2(x) - cos^2(x)  = -cos(x)

1 - 2cos^2(x)   = -cos(x)       add cos(x) to both sides and multiply through by - 1

2cos^2x - cos(x)  -  1  = 0     factor

(2 cos(x) + 1 )  (cos (x) - 1)  = 0       set each factor to 0

2cos(x) + 1 = 0    subtract 1 from both sides

2cos(x)   = -1     divide both sides by 2

cos(x)  = -1/2       and this is true when x = 120°   and x = 240°

For the other factor, we have

cos(x) - 1 = 0     add 1 to both sides

cos(x)   = 1     and this happens at 0°  in the indicated interval

So...the sum of the solutions is    [120 + 240 + 0 ]°  = 360°

Here's a graph of the intersection points :  https://www.desmos.com/calculator/c0oaxih2sy

Jennyara....

The solutions themselves have to occur within the given interval [ which these do ].....however.....the sum of such solutions might be ≥ 360°.....these happen to  sum to exactly to 360°