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Q.
The solution of the equation cos2x−2cosx=4sinx+4sin2x(0≤x≤π) is
Inverse Trigonometric Functions
Solution:
We have, cos2x−2cosx=4sinx+4sin2x cos2x−4sin2x=4sinx+2cosx (cosx+2sinx)(cosx−2sinx−2)=0 tanx=−12,cosx−2sinx=2 1−tan2x21+tan2x2−2⋅2tanx21+tan2x2=2 1−tan2x2−4tanx2=2+2tan2x2⇒3tan2x2+4tanx2+1=0 (3tanx2+1)(tanx2+1)=0⇒tanx2=−13 or −1 ⇒x2=tan−1(13) or −π4⇒x=−2tan−1(13),−π2]