Question

решите тригонометрическое уровнение
$\sin(x ) + 2 \cos(x) = 1$
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Answer 1

$\sin(x) + 2\cos(x) = 1$

$\sin(x) + 2\cos(x) \equiv \sqrt{5}\cdot(\sin(x)\cdot\frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}}\cdot\cos(x)) \equiv$

$\equiv \sqrt{5}\cdot(\sin(x)\cdot\cos(\phi) + \sin(\phi)\cdot\cos(x)) \equiv$

$\equiv \sqrt{5}\cdot\sin(x+\phi)$

$\phi = \mathrm{arcsin}(\frac{2}{\sqrt{5}})$

$\sqrt{5}\cdot\sin(x+\phi) = 1$

$\sin(x+\phi) = \frac{1}{\sqrt{5}}$

$x+\phi = (-1)^n\cdot\mathrm{arcsin}(\frac{1}{\sqrt{5}}) + \pi\cdot m$

$x = -\phi + (-1)^n\cdot\mathrm{arcsin}(\frac{1}{\sqrt{5}}) + \pi\cdot m$

$x = -\mathrm{arcsin}(\frac{2}{\sqrt{5}}) + (-1)^n\cdot\mathrm{arcsin}(\frac{1}{\sqrt{5}}) + \pi\cdot m$

$m\in\mathbb{Z}$