Given: 4sin2x−4cos3x+9−4cosx=0
⇒4−4cos2x−4cos3x+9−4cosx=0
⇒4cos3x+4cos2x+4cosx−13=0
⇒4cos3x+4cos2x+4cosx=13
We know that, −1≤cosx≤1
So, the maximum value of LHS is (4+4+4)=12
But, RHS is 13.
So, no solutions are possible for the given equation.
The number of solutions of the equation 4sin2x−4cos3x+9−4cosx=0;x∈[−2π,2π] is:
Held on 1 Feb 2024 · Verified 6 Jul 2026.
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