2sin^2x-root 3 sinx=0

Question

find all solutions in the interval [0, 2pi)

Answer 1

Let Z = sin x

 

2Z^2 - sqrt(3)*z = 0

 

Z ( 2Z - sqrt(3)) = 0

 

Z = 0     or 2*Z - sqrt(3) = 0

 

Z = 0   or Z = sqrt(3) /2

 

sin x = 0  or sin x = sqrt(3)/2

 

In the interval [0, 2*pi]

The sine function is zero ( or the inverse-sine at zero) are the angles 0 and pi,

 so those are two solutions.

 

The sine function is positive in quadrants 1 and 2 ,  (All Students take calculus or add sugar to coffee)

and we want flavors of 60 degrees where the sine function is sqrt(3)/2 = rad3/2 ( or inverse-sine(sqrt(3)/2))

so there are two more solutions of 60 degrees and 120 degrees, or in radians, pi/3 and 2*pi/3

 

The four solutions are zer0, pi/3, 2*pi/3, and pi