Question

1)вычислите cos(-arcsin sqrt(2)/2)= 2)найдите нули функции y=sin(x/2-п/2)= 3)решите уравнения: 1. 2cosx-sqrt(3)=0 2. sin^2 x + 2sinx +1 =0

Answer 1

$cos(-arcsin\sqrt{2}/2)=cos(-\pi/4)=cos(\pi/4)=\sqrt{2}/2\\\\y=sin(x/2-\pi/2)\\sin(x/2-\pi/2)=0\\x/2-\pi/2=\pi*n, n\in Z\\x/2=\pi/2+\pi*n, n\in Z|*2\\x=\pi+2\pi*n,n\in Z\\\\2cosx-\sqrt{3}=0\\cosx=\sqrt{3}/2\\x=+-\pi/6+2\pi*n,n\in Z\\\\sin^2x+2sinx+1=0\\t=sinx\\t^2+2t+1=0\\(t+1)^2=0\\t_{1,2}=-1\\sinx=-1\\x=3\pi/2+2\pi*k,k\in Z$