How do you solve $5sin2x-6sinx=0$ for $0<=x<=2pi$ ?

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Answer 1 · 2016-11-06T23:30:09

Solutions:
$x=0, x=pi,x=2pi,xapprox.927295,xapprox5.35589$

$5sin2x-6sinx=0$

First, use the double angle identity to simplify the $sin2x$ :
$5(2sinxcosx)-6sinx=0$

From here, we'll simplify and factor:
$10sinxcosx-6sinx=0$
$(2)(sinx)(5cosx-3)=0$

Now set each factor (that includes x) equal to zero:
$sinx=0$
$x=0$
$x=pi$
$x=2pi$

$5cosx-3=0$
$xapprox.927295$
$xapprox5.35589$

How do you solve 5sin2x-6sinx=05sin2x-6sinx=0 for 0<=x<=2pi0<=x<=2pi?