Question

How do i find the exact solution equation in the domain -2π ≤ x ≤ 2π. for 2 sin x + √3 = 0 ?

Answer 1

`x=(4pi)/3,x=(5pi)/3`

Explanation:

`"Isolate "sinx" by rearranging terms"`

`rArrsinx=-(sqrt3)/2`

Since the ratio is negative this informs us that x must be in the 3rd or 4th quadrants. Sine is positive in 1st/2nd quadrants.

We require the `color(blue)"related acute angle"` which is found by 'dropping' the negative from the ratio.

`rArrx=sin^-1(sqrt3/2)`

`rArrx=pi/3larrcolor(blue)"related acute angle"`

Find the relative values for x in the 3rd/4th quadrants.

`• color(red)" 3rd quadrant-positive direction-anti-clockwise"`

`x=pi+pi/3=(4pi)/3`

`"Eqivalent angle in 3rd quadrant- negative direction - clockwise"`

`x=(4pi)/3-2pi=-(2pi)/3`

`• color(red)" 4th quadrant - positive direction"`

`x=2pi-pi/3=(5pi)/3`

`"Eqivalent angle in 4th quadrant - negative direction"`

`x=(5pi)/3-2pi=-pi/3`

Solutions : `x=(4pi)/3harrx=-(2pi)/3`

`x=(5pi)/3harrx=-pi/3`