Question

2 sin x>1
Корень из 2 cosx <1
sin( x-П/6)>=-1/2

Answer 1

2sinx>1
sinx>1/2
π/6+2πn<x<(π-π/6)+2πn; n-
π/6+2πn<x<5π/6 +2πn; n-целое

сosx<1/√2
arccos(1/√2)+2πn<x<(2π-arccos(1/√2))+2πn; n-
π/4+2πn<x<7π/4 +2πn, n-celoe

sin(x-π/6)>-1/2
-π/6+2πk<x-π/6<(π+π/6)+2πK, K-
-π/6+π/6+2πK<X<7π/6+π/6+2πK, K-
2πK<X<4π/3+2πK, K-CELOE