Solve $4 {sin}^{2} (x) - 3 = 0$ $4sin^2(x) -3 = 0$ for $[0, 2 π)$ $[0,2pi)$ ????

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Answer 1 · 2017-10-06T20:37:42

$\frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$ $pi/3 , (2pi)/3 , (4pi)/3, (5pi)/3$

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

Then:

$4 z^{2} - 3 = 0$ $4z^2 - 3 =0$

$z^{2} = \frac{3}{4} \Rightarrow z = \frac{\sqrt{3}}{2} \Rightarrow sin (x) = \frac{\sqrt{3}}{2}$ $z^2 = 3/4=> z= sqrt(3)/2=> sin(x)=sqrt(3)/2$

${sin}^{- 1} sin (x) = {sin}^{- 1} (\frac{\sqrt{3}}{2}) \Rightarrow x = \frac{π}{3}$ $sin^-1sin(x)= sin^-1(sqrt(3)/2)=>x= color(blue)(pi/3)$

$π - \frac{π}{3} = \frac{2 π}{3}$ $pi - pi/3= color(blue)((2pi)/3)$

$2 π - \frac{2 π}{3} = \frac{4 π}{3}$ $2pi - (2pi)/3 = color(blue)((4pi)/3)$

$2 π - \frac{π}{3} = \frac{5 π}{3}$ $2pi - pi/3 = color(blue)((5pi)/3)$

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