Solve 4sin^2(x) -3 = 04sin2(x)3=0 for [0,2pi)[0,2π)????

1 Answer
Oct 6, 2017

pi/3 , (2pi)/3 , (4pi)/3, (5pi)/3π3,2π3,4π3,5π3

Explanation:

Let z = sin(x)z=sin(x)

Then:

4z^2 - 3 =04z23=0

z^2 = 3/4=> z= sqrt(3)/2=> sin(x)=sqrt(3)/2z2=34z=32sin(x)=32

sin^-1sin(x)= sin^-1(sqrt(3)/2)=>x= color(blue)(pi/3)sin1sin(x)=sin1(32)x=π3

pi - pi/3= color(blue)((2pi)/3)ππ3=2π3

2pi - (2pi)/3 = color(blue)((4pi)/3)2π2π3=4π3

2pi - pi/3 = color(blue)((5pi)/3)2ππ3=5π3