Not getting this one. In the interval 0deg<=A< 360deg, solve for all values of A in the equation cos2A=-3sinA -1Thought I asked yesterday but my question didn't make it through!
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.