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Questions  

Number of solutions of the equation 4cos22x+cos2x+1 +tanx(tanx23)=0 in [0,2π] is

a
0
b
1
c
2
d
3

detailed solution

Correct option is C

4cos2⁡2x+4cos⁡2x+1+tan2⁡x−23tan⁡x+3=0⇒ (2cos⁡2x+1)2+(tan⁡x−3)2=0⇒ cos⁡2x=−12 and tan⁡x=3⇒ 2cos2⁡x−1=−12 and tan⁡x=3⇒ cos⁡x=±12 and tan⁡x=3⇒ x=π3,4π3Thus, number of solutions of equation is 2.

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