How do you solve 3sinA=1+cos2A?

1 Answer
Jun 1, 2016

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

Explanation:

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#