How do I solve for all real values of x in this equation 2 cos² x = 3 sin x ?

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How do I solve for all real values of x in this equation 2 cos² x = 3 sin x ?

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Answer 1 · 2018-02-22T07:40:39

$x = \frac{π}{6} + 2 k π$ $x=pi/6+2kpi$
$x = \frac{5 π}{6} + 2 k π$ $x=(5pi)/6+2kpi$

Explanation:

$2 {cos}^{2} x = 3 sin x$ $2cos^2x=3sinx$
$2 \cdot (1 - {sin}^{2} x) = 3 sin x$ $2*(1-sin^2x)=3sinx$
$2 - 2 {sin}^{2} x = 3 sin x$ $2-2sin^2x=3sinx$
$2 {sin}^{2} x + 3 sin x - 2 = 0$ $2sin^2x+3sinx-2=0$

$sqrt(∆)=sqrt(25)=5$
$t_1=(-3-5)/4=-2$
$t_2=(-3+5)/4=1/2$

$sinx=1/2$

$x=pi/6+2kpi$
$x=(5pi)/6+2kpi$

k is real

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How do I solve for all real values of x in this equation 2 cos² x = 3 sin x ?

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