sinx+cosx>0如何解

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sinx+cosx>0如何解
sinx+cosx>0如何解

sinx+cosx>0如何解
sinx+cosx=√2sin(x-π/4)>0
2kπ

sinx+cosx>0
两边同乘以(√2/2)
(√2/2)sinx+(√2/2)cosx>0
sinxcosπ/4+cosxsinπ/4>0
sin(x+π/4)>0
2kπ2kπ-π/4

sinx+cosx>0,得到:√2sin(x+π/4)>0
则:sin(x+π/4)>0
2kπ得:2kπ-π/4

sinx+cosx>0
√2(sinxcosπ/4+cosxsinπ/4)>0
sin(x+π/4)>0
2kπ2kπ-π/4

sin(x+Pi/4)>0
(2k+1)Pi>x+Pi/4>2kPi
(2k+3/4)Pi>x>(2k-1/4)Pi