Question

2 sin квадрат x-cos x=1

Answer 1

Ответ:

$2sin^{2}x -cosx=1\2-2cos^{2} x-cosx=1\2cos^{2} x+cosx-1=0\t=cosx(|t|leq 1)\2t^{2} +t-1=0\D=1+8=9\t_{1} =frac{-1+3}{4} =frac{1}{2}\t_{2}=frac{-1-3}{4}=-1\ cosx=-1\x_{1} =pi +2pi k\cosx=frac{1}{2}\ x_{2} =frac{pi }{3}+2pi k\x_{3} =frac{5pi }{3}+2pi k$

Объяснение: