Solve 2sin(2x) - cosx = 0 for (0,2pi) ?

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Answer 1 · 2018-04-11T03:12:27

#x=pi/2, (3pi)/2, x approx 0.25#

Recall the identity #sin(2x)=2sinxcosx.# We may then rewrite as

#2(2)sinxcosx-cosx=0#

#4sinxcosx-cosx=0#

Factor out the cosine, as all terms share a cosine in common.

#cosx(4sinx-1)=0#

We must now solve #cosx=0, 4sinx-1=0#.

#cosx=0 -> x=pi/2, (3pi)/2# on #(0, 2pi)#

#4sinx-1=0#

#4sinx=1#

#sinx=1/4#

This one has no answer that could be found just from looking at the unit circle, I believe. However, using a calculator, #x=arcsin(1/4)approx0.25#