Q.

If θ[2π,2π], then the number of solutions of 22cos2θ+(26)cosθ3=0, is equal to :          [2025]

1 8  
2 12  
3 10  
4 6  

Ans.

(1)

We have, 22cos2θ+2cosθ6cosθ3=0

 (2cosθ3)(2cosθ+1)=0  cosθ=[32,12]

 11π6,π6,π6,11π6,5π4,3π4,5π4,3π4

Hence, number of solutions = 8.