# How do you find the general solutions for sqrt{3}(sin(x)) + cos(x) = 1?

Aug 12, 2015

$x = \left(6 n + {\left(- 1\right)}^{n} - 1\right) \frac{\pi}{6} , \text{ } n \in \mathbb{Z}$

#### Explanation:

$R = \sqrt{{a}^{2} + {b}^{2}}$
$a \sin x + b \cos x = R \left(\frac{a}{R} \sin x + \frac{b}{R} \cos x\right) = R \sin \left(x + \alpha\right)$
where $\frac{a}{R} = \cos \alpha$ and $\frac{b}{R} = \sin \alpha$ such that $\alpha = \arctan \left(\frac{b}{a}\right)$

$\sqrt{3} \sin x + \cos x = 2 \sin \left(x + \frac{\pi}{6}\right) = 1$
$\sin \left(x + \frac{\pi}{6}\right) = \frac{1}{2}$

$x + \frac{\pi}{6} = n \pi + {\left(- 1\right)}^{n} \frac{\pi}{6} , \text{ } n \in \mathbb{Z}$
$x = \left(6 n + {\left(- 1\right)}^{n} - 1\right) \frac{\pi}{6} , \text{ } n \in \mathbb{Z}$