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

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mason m Share
May 9, 2016

Answer:

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

Explanation:

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#

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