[2cos³x－sin²(360°－x)＋2sin(90°+x)+1]÷[2+2cos²(180°+x)+cos(－x)]
=[2cos³x-sin²x+2cosx+1]/[2+cos²x+cosx]
=[2cos³x-1+cos²x+2cosx+1]/[2+cos²x+cosx]
=cosx[2+cos²x+cosx]/[2+cos²x+cosx]
=cosx
化简：[2cos³x－sin²(360°－x)＋2sin(90°+x)+1]÷[2+2cos&sup
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