Question #148497

3cos(x-30)=cos0

Expert's answer

3⋅cos⁡(x−30°)=cos⁡0°3⋅cos⁡(x−30°)=1cos⁡(x−30°)=13cos⁡(π6−x)=13π6−x=arccos⁡(13)+2πn1,n1∈Z π6−x=−arccos⁡(13)+2πn2,n2∈Z−x=arccos⁡(13)−π6+2πn1,n1∈Z −x=−arccos⁡(13)−π6+2πn2,n2∈Zx=π6−arccos⁡13−2πn1,n1∈Z x=π6+arccos⁡13−2πn2,n2∈ZAnswer:x=π6−arccos⁡13−2πn1,n1∈Z x=π6+arccos⁡13−2πn2,n2∈Z3 \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


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