Question

ПОМОГИТЕ РЕШИТЬ ТРИГОНОМЕТРИЧЕСКОЕ УРАВНЕНИЕ ПОЖАЛУЙСТА!!!!

Answer 1

$sqrt{10}cosx-sqrt{4cosx-cos2x}=0\sqrt{10}cosx=sqrt{4cosx-cos2x}\cosx geq0\ (sqrt{10}cosx) ^{2}=(sqrt{4cosx-cos2x}) ^{2}\10cos^{2}x=4cosx-cos2x\10cos^{2}x=4cosx-(2cos^{2}x-1 )\10cos^{2}x=4cosx-2cos^{2}x+1\10cos^{2}x+2cos^{2}x-4cosx-1=0\12cos^{2}x-4cosx-1=0\cosx=a\12a^{2}-4a-1=0\D=16+48=64\a_{1}=frac{4+8}{24}=frac{12}{24}=frac{1}{2}\a_{2}=frac{4-8}{24}=-frac{4}{24}=-frac{1}{6}\cosx=frac{1}{2}\x= pm arccos(frac{1}{2} ) + 2pi n = pm frac{pi}{3} + 2pi n$