Hello from MrBillDoesMath!
Answer:
See Discussion below
Discussion:
2 (cos@)^2 + sqrt(3 cos@) = 0 => subtract sqrt(..) from both sides
2 (cos@)^2 = - sqrt(3 cos@) => square both sides
4 (cos2)^4 = 3 (cos@)^2 => subtract 3 (cos@)^2 from both sides
4(cos@)^4 - 3(cos@)^2 =0 => factor (cos@)^2 from each term
(cos@)^2 ( 4 (cos@)^2 -3 = 0) => A*B = implies A or B or both are 0
cos@ = 0 or (*)
cos@ = +\-sqrt(3/4)
cos@ = +\-sqrt(3)/2 (**)
Solution of (*) is @ = Pi/2 + Pi*n, n = { 0, +\1, +\- 2, +\- 3,...)
Solution of (**) is @ = Pi/6 + Pi*n , n = { 0, +\1, +\- 2, +\- 3,...)
General solution is union of the values in (*) and (**)
Thank you,
MrB
