Question

корень из 3 на sinx-cosx=2

Answer 1

cosx/4-sqrt(3)/4*sinx=-1/2
Пусть cosa=1/4, sina=sqrt(3)/4
cos(x+a)=-1/2
x+a=+-2pi/3+2pik
x=+-2pi/3-arccos(1/4)+2pik

Answer 2

$\mathtt{\sqrt3sinx-cosx=2}\\\mathtt{\sqrt3sinx-cosx-2=0}\\\mathtt{ \dfrac{\sqrt3}{2}sinx- \dfrac{1}{2}cosx=1}\\\mathtt{cos \dfrac{\pi}{6} sinx-sin \dfrac{\pi}{6} cosx=1}\\\mathtt{-(sin \dfrac{\pi}{6}-x)=1}\\\mathtt{sin( \dfrac{\pi}{6}-x)=-1}\\\mathtt{ \dfrac{\pi}{6}-x=- \dfrac{\pi}{2}+2\pi k,k\in\mathbb{Z}}\\\mathtt{x= \dfrac{\pi}{2}- \dfrac{\pi}{6}-2\pi k,k\in\mathbb{Z}}\\\boxed{\mathtt{x= \dfrac{\pi}{3}-2\pi k,k\in\mathbb{Z}}}$