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

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Answer 1 · 2017-10-06T20:37:42

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

Let #z = sin(x)#

Then:

#4z^2 - 3 =0#

#z^2 = 3/4=> z= sqrt(3)/2=> sin(x)=sqrt(3)/2#

#sin^-1sin(x)= sin^-1(sqrt(3)/2)=>x= color(blue)(pi/3)#

#pi - pi/3= color(blue)((2pi)/3)#

#2pi - (2pi)/3 = color(blue)((4pi)/3)#

#2pi - pi/3 = color(blue)((5pi)/3)#