The above answer is right but it is not correct, since we have to find the general solution of this equation.
We have to find A in `0lt=Alt=360` degrees.
We can write `cos(2A)` as,
`cos(2A) = 1-2sin^2(A)`
`1-2sin^2(A) = -3sin(A)-1`
`2sin^2(A)-3sin(A)-2 = 0`
`(2sin(A)+1)(sin(A)-2) = 0`
`2sin(A)+1 = 0` and `sin(A)-2 = 0`
We know `sin(A)` is in the region of `-1lt=sin(A)lt=+1` , therefore,
`sin(A) != 2`
The only solution is `2sin(A)+1 =0` and this gives,
`sin(A) = -1/2`
The primary solution is, `A =-30` degrees.
Let's find the general solution of A in the given range. The general solution for sine is given as,
`A = n(180)+(-1)^n(-30)` where `n in Z` .
For `n = 0` ,
`A = -30,` which is not in the range.
`n= 1, A = 180 - (-30) = 210.`
`n =2, A = 360+(-30) = 330.`
Therefore the required solutions are A = 210 and A =330 degrees.