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√(sin2x)=√(sinx) sin2x=sinx 2sinxcosx=sinx (i)若sinx=0 顯然符合 那麼x=kπ(k∈Z) (ii)若sinx≠0 則2cosx=1 即cosx=1/2 所以sinx=√(1-(1/2)^2)=√3/2(為什麼取正的,是因為根號內的是非負的) 所以x在第一象限角 那麼x=2kπ+π/3(k∈Z) 綜上,解集為{x|x=kπ或2kπ+π/3(k∈Z)}
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