已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1 求使f(x)大于等于零成立的x取值范围.我知道答案,是:f(x)=2sin(x+π/6)-1因为:f(x)≥0所以:2sin(x+π/6)-1≥0即:sin(x+π/6)≥1/2有:2kπ+5π/6≥x+π/6≥2k

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已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1求使f(x)大于等于零成立的x取值范围.我知道答案,是:f(x)=2sin(x+π/6)-1因为:f(x)≥0所以

已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1 求使f(x)大于等于零成立的x取值范围.我知道答案,是:f(x)=2sin(x+π/6)-1因为:f(x)≥0所以:2sin(x+π/6)-1≥0即:sin(x+π/6)≥1/2有:2kπ+5π/6≥x+π/6≥2k
已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1 求使f(x)大于等于零成立的x取值范围.
我知道答案,是:f(x)=2sin(x+π/6)-1
因为:f(x)≥0所以:2sin(x+π/6)-1≥0即:sin(x+π/6)≥1/2有:2kπ+5π/6≥x+π/6≥2kπ+π/6,k∈Z解得:2kπ+2π/3≥x≥2kπ,k∈Z
但是为什么要x

已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1 求使f(x)大于等于零成立的x取值范围.我知道答案,是:f(x)=2sin(x+π/6)-1因为:f(x)≥0所以:2sin(x+π/6)-1≥0即:sin(x+π/6)≥1/2有:2kπ+5π/6≥x+π/6≥2k

就看【-π,π),这个周期,正弦值不小于0.5的范围就是【π/5,5π/6】,不是π/2

f(x)=2sin(x+π/6)-1