2cos^2thetw plus sq root of 3 cos theta equals zero

Hello from MrBillDoesMath!

See Discussion below

Discussion:

2 ([email protected])^2  +  sqrt(3 [email protected]) = 0        => subtract sqrt(..) from both sides

2 ([email protected])^2 = - sqrt(3 [email protected])              => square both sides

4 (cos2)^4 = 3 ([email protected])^2                    => subtract 3 ([email protected])^2   from both sides

4([email protected])^4 - 3([email protected])^2 =0                 => factor ([email protected])^2 from each term

([email protected])^2 ( 4 ([email protected])^2 -3 = 0)            => A*B = implies A or B or both are 0

[email protected] = 0   or             (*)

[email protected] = +\-sqrt(3/4)

[email protected] = +\-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

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11 -1> 1 -1+ a

-12 > a