Question

1) 3sin(π/2-π)-√3=0
2) 3ctg(π/2+x)-√3=0

Answer 1

1е:
3*sin(π/2 + π) - $sqrt{3}$ = 0
3*cos(π) = $sqrt{3}$
cos(π) = $1/sqrt{3}$
-1 = $1/sqrt{3}$, неверно

решений нет

2е:
3*ctg(π/2 + x) - $sqrt{3}$ = 0
-3*tg(x) = $sqrt{3}$
tg(x) = $-1/sqrt{3}$

x = 5π/6 + πn, n∈Z