# How do you solve sin(x+60°) + cos(x+30°) = 1/2 in the range 0° < x < 360°?

May 2, 2015

$x = {73.22}^{\circ}$ and $x = 360 - 73.22 = {286.78}^{\circ}$

#### Explanation:

Use the trig identities:
$\sin \left(a + b\right) = \sin a . \cos b + \sin b . \cos a$
$\cos \left(a + b\right) = \cos a . \cos b - \sin a . \sin b$

$\sin \left(x + 60\right) = \left(\frac{1}{2}\right) . \sin x + \left(\frac{\sqrt{3}}{2}\right) . \cos x$

$\cos \left(x + 30\right) = \cos x . \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right) . \sin x$

Adding the 2 expressions, we get:

$\left(\sqrt{3}\right) . \cos x = \frac{1}{2} \to \cos x = \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{6}$

->$x = {73.22}^{\circ}$ and $x = 360 - 73.22 = {286.78}^{\circ}$