How do you solve $3 sin A = 1 + cos 2 A$ $3sinA=1+cos2A$ ?

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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,
$quad :\quad A=\frac{\pi }{6}+2\pi n,\:A=\frac{5\pi }{6}+2\pi n$ #

How do you solve 3sinA=1+cos2A3sinA=1+cos2A3sinA=1+cos2A?