2 câu trả lời
cos2x+1cos2x=2(cosx−1cosx)+1DK:cosx≠0⇔x≠π2+kπDatt=cosx−1cosx⇒t2=cos2x+1cos2x−2⇒cos2x+1cos2x=t2+2PT:t2+2=2t+1⇔t2−2t+1=0⇔t=1⇒cosx−1cosx=1⇔cos2x−cosx−1=0⇔[cosx=1+√52(loai)cosx=1−√52⇔x=±arccos1−√52+k2π(k∈Z)
cos2x+1cos2x=2(cosx−1cosx)+1DK:cosx≠0⇔x≠π2+kπDatt=cosx−1cosx⇒t2=cos2x+1cos2x−2⇒cos2x+1cos2x=t2+2PT:t2+2=2t+1⇔t2−2t+1=0⇔t=1⇒cosx−1cosx=1⇔cos2x−cosx−1=0⇔[cosx=1+√52(loai)cosx=1−√52⇔x=±arccos1−√52+k2π(k∈Z)