please-help-me-to-solve-it-in-R-3cosx-3-sinx-6-0-

Question Number 75763 by mathocean1 last updated on 16/Dec/19

please help me to solve it in R

3 cosx - \sqrt{3} sinx + \sqrt{6} = 0

Answered by Ajao yinka last updated on 16/Dec/19

Answered by $@ty@m123 last updated on 17/Dec/19

D i v i d e b y 2 \sqrt{3},

\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x + \frac{1}{\sqrt{2}} = 0

\sin \frac{π}{3} \cos x - \cos \frac{π}{3} \sin x = - \frac{1}{\sqrt{2}}

\sin (\frac{π}{3} - x) = - \frac{1}{\sqrt{2}}

\sin (x - \frac{π}{3}) = \frac{1}{\sqrt{2}}

\sin (x - \frac{π}{3}) = \sin \frac{π}{4}

x - \frac{π}{3} = n π + {(- 1)}^{n} \frac{π}{4}, n \in Z

x = n π + {(- 1)}^{n} \frac{π}{4} + \frac{π}{3}