How do you solve $sin2x-cosx=0$ over the interval 0 to 2pi?

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Answer 1 · 2016-05-09T21:02:27

$x=pi/6,pi/2,(5pi)/6,(3pi)/2$

Use the identity $sin2x=2sinxcosx$ .

$2sinxcosx-cosx=0$

Factor a $cosx$ term on the left hand side.

$cosx(2sinx-1)=0$

Set both of these terms equal to $0$ .

$cosx=0" "=>" "x=pi/2,(3pi)/2$

$2sinx-1=0" "=>" "sinx=1/2" "=>" "x=pi/6,(5pi)/6$

How do you solve sin2x-cosx=0 sin2x-cosx=0 over the interval 0 to 2pi?