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:

$4z^2 - 3 =0$

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

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

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

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

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

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