Question

Решать Уравнение 6sin^2 x- 5sinx+1=0

Answer 1

$6sin^2x-5sinx+1=0\\ sinx=t, t \in [-1;1]\\ 6t^2-5t+1=0\\ D=(-5)^2-4\cdot6\cdot1=25-24=1\\ t_1= \frac{5+1}{12}= \frac{6}{12} = \frac{1}{2} \\ t_2= \frac{5-1}{12}= \frac{4}{12} = \frac{1}{3} \\ ====================\\ sinx= \frac{1}{2} \\ x=(-1)^k arcsin \frac{1}{2}+ \pi k, k \in Z\\ x=(-1)^k \frac{ \pi }{6}+ \pi k, k \in Z\\ ====================\\ sinx= \frac{1}{3} \\ x=(-1)^n arcsin \frac{1}{3}+ \pi n, n \in Z$