Sin x< - √2/2
(- 3п/4 + 2Пn; - п/4 + 2Пn)
решение во вложении
-----------------------------------------
=двенадцать двенадцатых =1
<span>1) (1+ ctg</span>β<span>)</span>²<span>+ (1 - ctg</span>β<span> )</span>²=1+2ctgβ+ctg²β+1-2ctgβ+ctg²β=
=2+2ctg²β=2(1+ctg²β)=1/sin²β;<span>
2) </span>
<span>
3) </span>
<span>
4) </span>