Answer to Question #148497 in Trigonometry for Sajid

Question

3cos(x-30)=cos0

Answer 1

$3 \cdot \cos (x-30\degree) = \cos 0\degree \\ 3 \cdot \cos (x-30\degree) = 1\\ \cos (x-30\degree) = \frac{1}{3} \\ \cos (\frac{\pi}{6}-x) = \frac{1}{3} \\ \frac{\pi}{6}-x=\arccos(\frac{1}{3})+2\pi n_1, n_1\in Z\ \\ \frac{\pi}{6}-x=-\arccos(\frac{1}{3})+2\pi n_2, n_2\in Z \\ -x=\arccos(\frac{1}{3})-\frac{\pi}{6}+2\pi n_1, n_1\in Z\ \\ -x=-\arccos(\frac{1}{3})-\frac{\pi}{6}+2\pi n_2, n_2\in Z \\ x=\frac{\pi}{6}-\arccos\frac{1}{3}-2\pi n_1, n_1\in Z\ \\ x=\frac{\pi}{6}+\arccos\frac{1}{3}-2\pi n_2, n_2\in Z \\ Answer:\\ x=\frac{\pi}{6}-\arccos\frac{1}{3}-2\pi n_1, n_1\in Z\ \\ x=\frac{\pi}{6}+\arccos\frac{1}{3}-2\pi n_2, n_2\in Z$