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sin2x(cotx+tan2x)=4cos2x Điều kiện {sinx≠0cos2x≠0⇔{x≠kπx≠π4+kπ2,k∈Z Phương trình ⇔sin2xcotx+sin2x.tan2x=4cos2x ⇔2cos2x+sin22xcos2x=4cos2x ⇔1−cos22xcos2x=cos2x+1 ⇔1−cos22x=cos22x+cos2x ⇔2cos22x+cos2x−1=0 ⇔{cos2x=−1cos2x=12 ⇔{2x=π+2kπ2x=±π3+2kπ⇔{x=π2+kπx=±π6+kπ(k∈Z) (thỏa mãn)
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