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How do you find all the solutions between 0 and 2π for $4 {sin}^{2} x - 3 = 0$ $4sin^2x-3 = 0$ ?

1 Answer

Aug 17, 2017

$x \in {\frac{π}{3}, \frac{2 π}{3}, \frac{5 π}{3}, \frac{7 π}{3}}$ $x in {pi/3, (2pi)/3, (5pi)/3, (7pi)/3}$

Explanation:

If $4 {sin}^{2} x - 3 = 0$ $4sin^2x-3=0$
then
$XXX {sin}^{2} x = \frac{3}{4}$ $color(white)("XXX")sin^2x=3/4$
and
$XXX sin x = \pm \frac{\sqrt{3}}{2}$ $color(white)("XXX")sin x=+-sqrt(3)/2$

This is one of the standard reference angles $= \frac{π}{3}$ $= pi/3$

Between 0 and $2 π$ $2pi$ , the possibilities for this reference angle are
$XXX \frac{π}{3}, \frac{2 π}{3}, \frac{5 π}{3}, \frac{7 π}{3}$ $color(white)("XXX")pi/3,(2pi)/3, (5pi)/3, (7pi)/3$

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