题目内容
若x=
【答案】分析:把x=
代入方程2cos(x+α)=1,化简根据α∈(0,2π),确定函数值的范围,求出α.
解答:解:∵x=
是方程2cos(x+α)=1的解,
∴2cos(
+α)=1,即cos(
+α)=
.
又α∈(0,2π),∴
+α∈(
,
).
∴
+α=
.∴α=
.
故答案为:
点评:本题考查三角函数值的符号,三角函数的定义域,考查逻辑思维能力,计算能力,是基础题.

解答:解:∵x=

∴2cos(



又α∈(0,2π),∴



∴



故答案为:

点评:本题考查三角函数值的符号,三角函数的定义域,考查逻辑思维能力,计算能力,是基础题.

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