若cos(π/4+x)=3/5 17π/12
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若cos(π/4+x)=3/517π/12若cos(π/4+x)=3/517π/12若cos(π/4+x)=3/517π/12解析：若cos(π/4+x)=3/5,那么：sin2x=-cos(π/2+
若cos(π/4+x)=3/5 17π/12
若cos(π/4+x)=3/5 17π/12
若cos(π/4+x)=3/5 17π/12
解析：
若cos(π/4+x)=3/5,那么：
sin2x=-cos(π/2 +2x)
=-[2cos²(π/4+x) -1]
=-2cos²(π/4+x) +1
=-18/25 +1
=7/25
7/25
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若cos(π/4+x)=3/5 17π/12
若cos（π/4+x）=3/5,17/12π
若cos(π/4+x)=3/5,17π/12
若cos(π/4+x)=3/5,17π/12
若cos(π/4+x)=3/5 17π/12
若cos(π/4+x)=3/5 17π/12
若cos（π/4+x)=3/5,17/12π