Question

How do you solve `2sin(x)-3cot(x)-3csc(x)=0` for `x`?

Answer 1

`x=n360+-180, n360+-120`
`x=2n\pi+-\pi, 2n\pi+-(2\pi)/3`

Explanation:

The equation can be rewritten as `(2sinx)/1-(3cosx)/sinx-3/sinx=0`

If we multiply both sides by `sinx` we get `2sin^2x-3cosx-3=0`

Using the identity: `sin^2x+cos^2x-=1` We can find that `sin^2x-=1-cos^2x`

So, `2(1-cos^2x)-3cosx-3=0`

`2-2cos^2x-3cosx-3=-2cos^2x-3cosx-1=0`

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

By substituting `x` for `cos` we get: `2x^2+3x+1=0`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-3+sqrt(3^2-4(2*1)))/(2(2))=-1/2`

`x=(-3-sqrt(3^2-4(2*1)))/(2(2))=-1`

`cosx=-1 or -1/2`

`x=arccos(-1)=180`
`x=arccos(-1/2)=120`

However, `cos(360+180)=-1` and `cos(360-120)=-1/2`

`x=n360+-180, n360+-120`
`x=2n\pi+-\pi, 2n\pi+-(2\pi)/3`

Answer 2

For `x in [0,pi)`
`color(white)("XXX")x=(2pi)/3`

Explanation:

Note that neither `cot(x)` nor `csc(x)` are defined for `x=kpi, k in ZZ`

So the given equation
`color(white)("XXX")2sin(x)-3cot(x)-3csc(x)=0`
is not valid for `x=0`

Multiplying by `sin(x)`
`color(white)("XXX")2sin^2(x)-3cos(x)-3=0`

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

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

`color(white)("XXX")2cos^2(x)+3cos(x)+1=0`

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

#color(white)("XXX"){: ((2cos(x)+1)=0," or ",(cos(x)+1)=0), (rarr cos(x)=-1/2,,rarr cos(x)=-1), (rarr x = (2pi)/3,,rarr x=pi), (color(white)("xxxxx")"for " x in [0,pi],,color(white)("xxxxx")"for " x in [0,pi]), (,,"BUT the given expression is not defined for this value of "x) :}#