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

Jun 1, 2016

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

#### Explanation:

$3 \setminus \sin \left(A\right) = 1 + \setminus \cos \left(2 A\right)$

or,$- \cos \left(2 A\right) + 3 \sin \left(A\right) - 1 = 0$

using the identity,$\cos \left(2 x\right) = 1 - 2 {\sin}^{2} \left(x\right)$

$- \left(1 - 2 {\sin}^{2} \left(A\right)\right) - 1 + 3 \sin \left(A\right) = 0$

$- 2 + 2 {\sin}^{2} \left(A\right) + 3 \sin \left(A\right) = 0$

Let $\setminus \sin \left(A\right) = u$

$- 2 + 2 {u}^{2} + 3 u = 0$

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

substituting back,$\sin \left(A\right) = u$
$\sin \left(A\right) = \frac{1}{2} \mathmr{and} \sin \left(A\right) = - 2$

So,$\setminus \sin \left(A\right) = - 2$i.e none
and,$\sin \left(A\right) = \frac{1}{2}$ C

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