Question #77b55

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Answer 1 · 2016-11-19T17:26:12

#sinxcosx = 0 -> (sinxcosx)^2 = 0^2#

#sin^2xcos^2x = 0#

Use the identity #sin^2theta + cos^2theta =1-> cos^2theta = 1 -sin^2theta#.

#sin^2x(1 - sin^2x) = 0#

#sin^2x = 0 and sin^2x = 1#

#sinx = 0 and sinx = +-1#

#x = 0 + 2pin, pi + 2pin, pi/2 + 2pin, (3pi)/2 + 2pin#

Hopefully this helps!

Answer 2 · 2017-09-18T03:28:10

#x = pin, pi/2 + pin#

Alternatively, we have:

#2sinxcosx = sin(2x)#.

#1/2sin(2x) = 0#

#sin(2x) = 0#

#2x = 0, pi#

#x = 0 + pin, pi/2 + pin#

Hopefully this helps!