Given: #cos x + 1 = 0 " and " 4sin^2 x - 1 = 0#
These are two different equations, with different values of #x#.
To solve #cos x + 1 = 0:
#" "cos x = -1#
You can look at a unit trigonometric circle to find out that when #x = 180^o = pi, cos x = -1#:
unit trigonometric circle
Since the problem does not give a domain of angles, the answer requires all possible angles. Every time you rotate #360^o = 2 pi# you get the same value for the cosine.
Solution: #x = pi + 2n pi; n = ...-3, -2, -1, 0, 1, 2, ...#
To solve 4sin^2 x - 1 = 0:
This problem is the difference of squares: #(a^2-b^2) = (a + b)(a - b)#.
#4sin^2 x - 1 = (2sinx)^2 - 1^2 = (2sinx + 1) ( 2 sinx - 1) = 0#
#2 sinx + 1 = 0; " " 2 sin x - 1 = 0#
#2 sin x = -1; " " 2 sin x = 1#
#sin x = -1/2; " " sin x = 1/2#
If you look at the trig circle you will notice that the #sin x = -1/2# when #x = (7 pi)/6 " and " x = (11 pi)/6#
If you look at the trig circle you will notice that the #sin x = 1/2# when #x = (pi)/6 " and " x = (5 pi)/6#
Solution:
Since #x = (pi)/6# is #180^o# from #(7 pi)/6# we can write this answer as #x = (pi)/6 + n pi , n = ...-3, -2, -1, 0, 1, 2, ...#
Since #x = (5 pi)/6# is #180^o# from #(11 pi)/6# we can write this answer as #x = (5 pi)/6 + n pi , n = ...-3, -2, -1, 0, 1, 2, ...#
