Well, I think this method will be better.
Assume #cos beta = a#.
Now, The equation is,
#color(white)(xxx)a^2 + a - 2 = 0#
#rArr a^2 + (2 - 1)a - 2 = 0# [Break #1# as #2 - 1#]
#rArr a^2 + 2a - a -2 = 0# [Distributive Property]
#rArr a(a + 2) - 1(a + 2) = 0#
#rArr (a + 2)(a - 1)# [Grouping the like terms]
So, Either #a + 2 = 0 rArr a = -2#
Or, #a -1 = 0 rArr a = 1#
So, #cos beta = -2, 1#.
But cosine (#cos#) is a periodic function whose value ranges between #-1# and #1#. [#-1 <= cos beta <= 1#]
So, #cos beta != -2#.
And, #cos beta = 1#.
And We know, #cos (2npi) = 1#, where #n = ...-3, -2, -1, 0, 1, 2, 3....# so on.
So, for #n = 0#, #2npi = 0#,
That's why, #0^c# or #0^@# is a solution.
If #n = 1#, #2npi = 2pi#.
So, #2pi^c# or #360^@# is a solution too.
But, The interval #[0^@, 360^@)# is left-closed and right-open. So, #360^@# will not count.
So, #beta = 0^@#.
Hope this helps.
