How do you solve $2cos^2x=3sinx$ ?

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Answer 1 · 2016-11-22T19:41:30

$x=pi/6+2pin$
$x=5pi/6+2pin$

$2cos^2x=3sinx$

Rewrite $cos^2x$ as $1-sin^2x$ because of the identity $sin^2x+cos^2x=1$ :

$2(1-sin^2x)=3sinx$

Distribute the left side and let one side be zero:
$2-2sin^2x-3sinx=0$

$-2sin^2x-3sinx+2=0$

Factor:
$-2sin^2x-4sinx+sinx+2=0$
$-2sinx(sinx+2)+(sinx+2)=0$
$(-2sinx+1)(sinx+2)=0$

Use zero product rule:

$sinx+2cancel=0$
$sinxcancel=-2$

$-2sinx+1=0$
$-2sinx=-1$
$sinx=1/2$

$x=pi/6+2pin$
$x=5pi/6+2pin$

How do you solve 2cos^2x=3sinx2cos^2x=3sinx?