arctan1=_______________.
(2)若cosx=-
,x∈[0,π],则x为_______________.
arctan1=_______________.
(2)若cosx=-
,x∈[0,π],则x为_______________.
解
:(1)记x=arcsin(-∴-
=sinx.
∴x=-
.
记x=arctan1∈(-
,
),
∴1=tanx.∴x=
.
(2)当x∈[0,π]时cosx=-
,由反余弦定义,存在唯一的x=arccos(-
)为钝角,
即x=π-arccos
.
答案:
(1)-