How do you solve #Sin2x cosx - cos2x sinx = 1/2#?

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Answer 1 · 2016-02-21T06:02:25

#x=pi/6, (5pi)/6# or

#x=30^@, 150^@#

From sum and difference formulas

We have

#sin (x+y)=sin x cos y + cos x sin y#
#sin (x-y)=sin x cos y - cos x sin y#

#sin (2x-x)=sin 2x cos x - cos 2x sin x#

#sin x=sin 2x cos x - cos 2x sin x#

So

#sin x=1/2#

#x=sin^-1 (1/2)=pi/6, (5pi)/6#

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