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!

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thilina-g | College Teacher | (Level 1) Educator

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The above answer is right but it is not correct, since we have to find the general solution of this equation.

`cos(2A)=-3sin(A)-1`

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`

Therefore,

`2sin^2(A)-3sin(A)-2 = 0`

`(2sin(A)+1)(sin(A)-2) = 0`

This gives,

`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.

 

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