Question

Срочно!!!!!!!!!!!!1)(sinx-1/2)(sinx+1)=0 [0;2pi ]2)(cosx+1/2)(cosx-1)=0 [0;2pi]3)(cosx-√2/2)(sinx+√2/2)=0 [0;2pi]4)(1+cosx)(√2siСрочно!!!!!!!!!!!!1)(sinx-1/2)(sinx+1)=0 [0;2pi ]2)(cosx+1/2)(cosx-1)=0 [0;2pi]
3)(cosx-√2/2)(sinx+√2/2)=0 [0;2pi]
4)(1+cosx)(√2sinx-1)=0

Answer 1

Решение во вложении
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