![2sin\: 3x+cos \: 5x-\sqrt{3}\cdot sin\: 5x=0](https://tex.z-dn.net/?f=2sin%5C%3A%203x%2Bcos%20%5C%3A%205x-%5Csqrt%7B3%7D%5Ccdot%20sin%5C%3A%205x%3D0)
Будем применять метод вспомогательного аргумента (для 2-го и 3-го слагаемых):
![\displaystyle \sqrt{1^2+(-\sqrt{3})^2} \left(\frac{1}{\sqrt{1^2+(-\sqrt{3})^2}}\cdot cos \: 5x-\frac{\sqrt{3}}{\sqrt{1^2+(-\sqrt{3})^2}}\cdot sin\:5x \right) = \\ = 2\left(\frac{1}{2}\cdot cos\:5x-\frac{\sqrt{3}}{2}\cdot sin\:5x)=2\left(sin\frac{\pi}{6} \cdot cos\:5x-cos\frac{\pi}{6}\cdot sin\: 5x \right) =](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Csqrt%7B1%5E2%2B%28-%5Csqrt%7B3%7D%29%5E2%7D%20%5Cleft%28%5Cfrac%7B1%7D%7B%5Csqrt%7B1%5E2%2B%28-%5Csqrt%7B3%7D%29%5E2%7D%7D%5Ccdot%20cos%20%5C%3A%205x-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B1%5E2%2B%28-%5Csqrt%7B3%7D%29%5E2%7D%7D%5Ccdot%20sin%5C%3A5x%20%20%5Cright%29%20%3D%20%5C%5C%20%3D%202%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20cos%5C%3A5x-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5Ccdot%20sin%5C%3A5x%29%3D2%5Cleft%28sin%5Cfrac%7B%5Cpi%7D%7B6%7D%20%5Ccdot%20cos%5C%3A5x-cos%5Cfrac%7B%5Cpi%7D%7B6%7D%5Ccdot%20sin%5C%3A%205x%20%5Cright%29%20%3D)
![\displaystyle = 2sin\left(\frac{\pi}{6}-5x\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%202sin%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B6%7D-5x%5Cright%29)
После преобразования видно, что делать дальше.
![\displaystyle 2sin\:3x+2sin\left(\frac{\pi}{6}-5x\right)=0 \Rightarrow sin\:3x+sin\left(\frac{\pi}{6}-5x\right)=0 \Rightarrow \\ \Rightarrow 2sin\left(\frac{3x+\frac{\pi}{6}-5x}{2}\right)cos\left(\frac{3x-(\frac{\pi}{6}-5x)}{2}\right)=0 \Rightarrow \\ \Rightarrow 2sin\left(-x+\frac{\pi}{12} \right)cos\left(4x-\frac{\pi}{12}\right)=0 \Rightarrow](https://tex.z-dn.net/?f=%5Cdisplaystyle%202sin%5C%3A3x%2B2sin%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B6%7D-5x%5Cright%29%3D0%20%5CRightarrow%20sin%5C%3A3x%2Bsin%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B6%7D-5x%5Cright%29%3D0%20%5CRightarrow%20%5C%5C%20%5CRightarrow%202sin%5Cleft%28%5Cfrac%7B3x%2B%5Cfrac%7B%5Cpi%7D%7B6%7D-5x%7D%7B2%7D%5Cright%29cos%5Cleft%28%5Cfrac%7B3x-%28%5Cfrac%7B%5Cpi%7D%7B6%7D-5x%29%7D%7B2%7D%5Cright%29%3D0%20%5CRightarrow%20%5C%5C%20%5CRightarrow%202sin%5Cleft%28-x%2B%5Cfrac%7B%5Cpi%7D%7B12%7D%20%5Cright%29cos%5Cleft%284x-%5Cfrac%7B%5Cpi%7D%7B12%7D%5Cright%29%3D0%20%5CRightarrow)
![\displaystyle \Rightarrow \left [ {{sin\left(x-\frac{\pi}{12}\right)=0} \atop {cos\left(4x-\frac{\pi}{12}\right)=0}} \right. \Rightarrow \left [ {{x-\frac{\pi}{12}=\pi k,\: k\in \mathbb{Z}} \atop {4x-\frac{\pi}{12}=\frac{\pi}{2}+\pi n, \: n\in \mathbb{Z}}} \right. \Rightarrow](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5CRightarrow%20%5Cleft%20%5B%20%7B%7Bsin%5Cleft%28x-%5Cfrac%7B%5Cpi%7D%7B12%7D%5Cright%29%3D0%7D%20%5Catop%20%7Bcos%5Cleft%284x-%5Cfrac%7B%5Cpi%7D%7B12%7D%5Cright%29%3D0%7D%7D%20%5Cright.%20%5CRightarrow%20%5Cleft%20%5B%20%7B%7Bx-%5Cfrac%7B%5Cpi%7D%7B12%7D%3D%5Cpi%20k%2C%5C%3A%20k%5Cin%20%5Cmathbb%7BZ%7D%7D%20%5Catop%20%7B4x-%5Cfrac%7B%5Cpi%7D%7B12%7D%3D%5Cfrac%7B%5Cpi%7D%7B2%7D%2B%5Cpi%20n%2C%20%5C%3A%20n%5Cin%20%5Cmathbb%7BZ%7D%7D%7D%20%5Cright.%20%5CRightarrow)
![\displaystyle \Rightarrow \left [ {{x=\frac{\pi}{12}+ \pi k, \:k\in \mathbb{Z} } \atop {4x=\frac{7\pi}{12}+\pi n,\: n\in \mathbb{Z}}} \right. \Rightarrow \left [ {{x=\frac{\pi}{12}+ \pi k, \:k\in \mathbb{Z} } \atop {x=\frac{7\pi}{48}+\frac{\pi n}{4},\: n\in \mathbb{Z}}} \right.](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5CRightarrow%20%5Cleft%20%5B%20%7B%7Bx%3D%5Cfrac%7B%5Cpi%7D%7B12%7D%2B%20%5Cpi%20k%2C%20%5C%3Ak%5Cin%20%5Cmathbb%7BZ%7D%20%7D%20%5Catop%20%7B4x%3D%5Cfrac%7B7%5Cpi%7D%7B12%7D%2B%5Cpi%20n%2C%5C%3A%20n%5Cin%20%5Cmathbb%7BZ%7D%7D%7D%20%5Cright.%20%5CRightarrow%20%5Cleft%20%5B%20%7B%7Bx%3D%5Cfrac%7B%5Cpi%7D%7B12%7D%2B%20%5Cpi%20k%2C%20%5C%3Ak%5Cin%20%5Cmathbb%7BZ%7D%20%7D%20%5Catop%20%7Bx%3D%5Cfrac%7B7%5Cpi%7D%7B48%7D%2B%5Cfrac%7B%5Cpi%20n%7D%7B4%7D%2C%5C%3A%20n%5Cin%20%5Cmathbb%7BZ%7D%7D%7D%20%5Cright.)
Ответ: ![\boxed{x=\frac{\pi}{12}+ \pi k, \:k\in \mathbb{Z}; x=\frac{7\pi}{48}+\frac{\pi n}{4},\: n\in \mathbb{Z}}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D%5Cfrac%7B%5Cpi%7D%7B12%7D%2B%20%5Cpi%20k%2C%20%5C%3Ak%5Cin%20%5Cmathbb%7BZ%7D%3B%20x%3D%5Cfrac%7B7%5Cpi%7D%7B48%7D%2B%5Cfrac%7B%5Cpi%20n%7D%7B4%7D%2C%5C%3A%20n%5Cin%20%5Cmathbb%7BZ%7D%7D)
Формулы, которые использовались:
1. Введение дополнительного аргумента
![\displaystyle a\:cosx\pm b\:sinx=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cdot cosx \pm\frac{b}{\sqrt{a^2+b^2}}\cdot sinx \right) =\\=\sqrt{a^2+b^2} \left(sin\phi\cdot cosx\pm cos\phi\cdot sinx)=\sqrt{a^2+b^2}\cdot sin(\phi \pm x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%5C%3Acosx%5Cpm%20b%5C%3Asinx%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5Cleft%28%5Cfrac%7Ba%7D%7B%5Csqrt%7Ba%5E2%2Bb%5E2%7D%7D%5Ccdot%20cosx%20%5Cpm%5Cfrac%7Bb%7D%7B%5Csqrt%7Ba%5E2%2Bb%5E2%7D%7D%5Ccdot%20sinx%20%5Cright%29%20%3D%5C%5C%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cleft%28sin%5Cphi%5Ccdot%20cosx%5Cpm%20cos%5Cphi%5Ccdot%20sinx%29%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5Ccdot%20sin%28%5Cphi%20%5Cpm%20x%29)
![\displaystyle \phi = arcsin\frac{a}{\sqrt{a^2+b^2}}=arccos\frac{b}{\sqrt{a^2+b^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cphi%20%3D%20arcsin%5Cfrac%7Ba%7D%7B%5Csqrt%7Ba%5E2%2Bb%5E2%7D%7D%3Darccos%5Cfrac%7Bb%7D%7B%5Csqrt%7Ba%5E2%2Bb%5E2%7D%7D)
Кстати, если поменять косинус и синус местами, то все везде по ходу преобразований эти самые замены будут произведены и в итоговом синусе суммы внутри надо будет переставить аргументы местами (естественно, особое значение это играет при наличии знака "-")
2. Сумма/разность синусов:
![\displaystyle sin\:x\pm sin\:y=2\:sin\frac{x\pm y}{2}cos\frac{x\mp y}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20sin%5C%3Ax%5Cpm%20sin%5C%3Ay%3D2%5C%3Asin%5Cfrac%7Bx%5Cpm%20y%7D%7B2%7Dcos%5Cfrac%7Bx%5Cmp%20y%7D%7B2%7D)