cos2x=1 => x=0, т.к. cos 0 =1
cos(x-p/3)=1 => x=p/3, т.к. p/3-p/3=0, а cos 0 =1
sin(-x/2)=-1 => -sin(x/2)=-1 => sin(x/2)=1 => x=p, т.к. p/2 = 90°, a sin90°=1
sin(x+p/6)=1 => x+p/6 = p/2 => x=p/2-p/6 => x=p/3
cos(x/4)=0 => x/4 =1 , т.к. cos1=0 => x=4
Объяснение:
cos2x=1 => x=0, т.к. cos 0 =1
cos(x-p/3)=1 => x=p/3, т.к. p/3-p/3=0, а cos 0 =1
sin(-x/2)=-1 => -sin(x/2)=-1 => sin(x/2)=1 => x=p, т.к. p/2 = 90°, a sin90°=1
sin(x+p/6)=1 => x+p/6 = p/2 => x=p/2-p/6 => x=p/3
cos(x/4)=0 => x/4 =1 , т.к. cos1=0 => x=4
Объяснение: