Question

Question #55c95

Answer 1

In the range `x in [0,2pi]`
`color(white)("XXX")x=(2pi)/3 or (4pi)/3 or pi`

Explanation:

`2sin^2(x)-3cos(x)-3=0`

`rarr 2(1-cos^2(x))-3cos(x)-3=0color(white)("XXX")`(using Double Angle formula)

`rarr -2cos^2(x)-3cos(x)-1=0color(white)("XXX")`(simplifying)

`rarr (2cos(x)+1)(cos(x)+1)=0color(white)("XXX")`(factoring)

`rarr cos(x)=-1/2color(white)("XXX")`or`color(white)("XXX")cos(x)=-1`

For `x in [0,2pi]`
`color(white)("XXX")x= pi+-pi/3color(white)("XXXX")`or`color(white)("XXX")x=pi`

[Add `+n*2pi, n in ZZ` if you want not to restrict `x` to the range `[0,2pi]`

Answer 2

`2sin^2x-3cosx-3=0`

`=>2-2cos^2x-3cosx-3=0`

`=>2cos^2x+3cosx+1=0`

`=>2cos^2x+2cosx+cosx+1=0`

`=>2cosx(cosx+1)+1(cosx+1)=0`

`=>(2cosx+1)(cosx+1)=0`

So

`2cosx+1=0`

`=>cosx=-1/2=cos((2pi)/3)`

`:.x=2npi+-(2pi)/3 " where "n in ZZ`

Again

`cosx+1=0`

`=>cosx=-1=cos(pi)`

`:.x=2npi+pi=(2n+1)pi " where "n in ZZ`