Question

Найдите(в градусах) корень уравнения 4cos(48градусов-х)cos(42градусов+x)=√3 на промежутке(0;45)

Answer 1

Ответ:

$18^\circ$

Объяснение:

$4\cos (48^\circ - x)\cos (42^\circ + x) = \sqrt 3 .$

Воспользуемся формулой

$2\cos \alpha \cos \beta = \cos (\alpha - \beta ) + \cos (\alpha + \beta ).$

$4\cos (48^\circ - x)\cos (42^\circ + x) = 2\cos (48^\circ - x - 42^\circ - x) + 2\cos (48^\circ - x + 42^\circ + x) =\\\\= 2\cos (6^\circ - 2x) + 2\cos 90^\circ = 2\cos (6^\circ - 2x).$

Так как $\cos ( - \alpha ) = \cos \alpha,$ то

$2\cos (6^\circ - 2x) = 2\cos (2x - 6^\circ ) = 2\cos \left( {2x - \displaystyle\frac{\pi }{{30}}} \right).$

$2\cos \left( {2x - \displaystyle\frac{\pi }{{30}}} \right) = \sqrt 3 ;\\\\\cos \left( {2x - \displaystyle\frac{\pi }{{30}}} \right) = \displaystyle\frac{{\sqrt 3 }}{2};\\\\2x - \displaystyle\frac{\pi }{{30}} = \pm \arccos \displaystyle\frac{{\sqrt 3 }}{2} + 2\pi n;$

$2x = \displaystyle\frac{\pi }{{30}} \pm \displaystyle\frac{\pi }{6} + 2\pi n;\\\\x = \displaystyle\frac{\pi }{{60}} \pm \displaystyle\frac{\pi }{{12}} + \pi n;\\\\{x_1} = \displaystyle\frac{\pi }{{60}} + \displaystyle\frac{\pi }{{12}} + \pi n = \displaystyle\frac{\pi }{{10}} + \pi n,\\\\{x_2} = \displaystyle\frac{\pi }{{60}} - \displaystyle\frac{\pi }{{12}} + \pi k = - \displaystyle\frac{\pi }{{15}} + \pi k,\,\,k,\,\,n \in {\rm{Z}}.$

Промежутку $\left( {0;\,\,\displaystyle\frac{\pi }{4}} \right)$ принадлежит только один корень

$\displaystyle\frac{\pi }{{10}} = 18^\circ .$