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