the question asks to solve the problem in the interval 0
That's less than or equal to, btw.
-
cos^2x - 2cosx = 3
cos^2x - 2cosx - 3 = 0
(cos(x) -3)(cos(x)+1)=0
cos(x) cannot be 3 as it is suck between -1 and 1
so cos(x) must be -1 and thus x is pi
cos^2x - 2cosx - 3 = 0
(cos(x) -3)(cos(x)+1)=0
cos(x) cannot be 3 as it is suck between -1 and 1
so cos(x) must be -1 and thus x is pi
-
cos²x - 2cosx - 3 = 0
(cosx - 3)(cosx + 1) = 0
cosx = - 1
x = π
(cosx - 3)(cosx + 1) = 0
cosx = - 1
x = π