How do you solve #3sinA=1+cos2A#?

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Answer 1 · 2016-06-01T17:27:05

#3\sin (A)=1+\cos (2A)quad :\quad A=\frac{\pi }{6}+2\pi n,\:A=\frac{5\pi }{6}+2\pi n#

#3\sin (A)=1+\cos (2A)#

or,#-cos (2A)+3sin(A)-1=0#

using the identity,#cos (2x)=1-2sin ^2(x)#

#-(1-2sin ^2(A))-1+3sin (A)=0#

#-2+2sin ^2(A)+3sin (A)=0#

Let #\sin (A)=u#

#-2+2u^2+3u=0#

solving it,we get, #u=\frac{1}{2},\:u=-2#

substituting back,#sin(A)=u#
#sin(A)=1/2 and sin(A)=-2#

So,#\sin (A)=-2 #i.e none
and,#sin(A)=1/2# C

Finally,combining all the solutions,
#quad :\quad A=\frac{\pi }{6}+2\pi n,\:A=\frac{5\pi }{6}+2\pi n##