Find all values of x in [0^circ, 360^circ) to satisfy this equation? sin x + sqrt3= 3 sqrt3 cos x

Jun 3, 2018

59^@83; 278^@37

Explanation:

$\sin x + \sqrt{3} = 3 \sqrt{3} \cos x$
$\sin x - 3 \sqrt{3} \cos x = - \sqrt{3}$ (1)
Call $\tan t = \sin \frac{t}{\cos} t = 3 \sqrt{3} = 5.196$ -->
$t = {79}^{\circ} 10$, and cos t = 0.19
Equation (1) becomes:
$\sin x . \cos t - \sin t . \cos x = - \sqrt{3} \cos t = - 0.33$
$\sin \left(x - {79}^{\circ} 10\right) = - 0.33$
Calculator and unit circle give 2 solutions for (x - 79.10):

a. $x - 79.10 = - {19}^{\circ} 27$
$x = - 19.27 + 79.10 = {59}^{\circ} 83$
b. $x - 79.10 = 180 - \left(- 19.27\right) = {199}^{\circ} 27$
$x = 199.27 + 79.10 = {278}^{\circ} 37$
Check by calculator.
x = 59.83 --> sin x = 0.86 --> $\sin x + \sqrt{3} = 2.60$ -->
cos x = 0.50 -->$3 \sqrt{3.} \cos x = 2.60$. Proved.
x = 278.37 --> sin x = - 0.99 --> $\sin x + \sqrt{3} = 0.74$
cos x = 0.14 --> $3 \sqrt{3} \cos x = 0.75$. Proved