Question

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4 sin x + 16 cos x = корень 272
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Answer 1

$4sinx + 16cosx = \sqrt{272}\\\sqrt{272}(\frac{4}{\sqrt{272}}sinx + \frac{16}{\sqrt{272}}cosx) = \sqrt{272}\\ \phi = arccos\frac{4}{\sqrt{272}} = arcsin\frac{16}{\sqrt{272}} => cos\phi = \frac{4}{\sqrt{272}}, sin\phi = \frac{16}{\sqrt{272}} =>\\ cos\phi *sinx + sin\phi * cosx = 1\\sin(x+\phi) = 1\\x + \phi = \frac{\pi}{2} + 2\pi n, n \in Z\\ x = \frac{\pi}{2} - \phi + 2\pi n, n \in Z => x = \frac{\pi}{2} - arccos\frac{4}{\sqrt{272}}+ 2\pi n, n \in Z$

$Answer: x = \frac{\pi}{2} - arccos\frac{4}{\sqrt{272}}+ 2\pi n, n \in Z$