In the interval  the equation  has

The correct answer is: a unique solution

Here we have to find it which type of solution it has.
Firstly, we have given,

We know if  then we can take antilog, b = c^a
So, we can write,
Cos² θ = sin² θ
1 – 2 sin² θ = sin² θ [ since, cos² θ = 1 – 2 sin² θ ]
3sin² θ = 1
sin² θ =
sin θ = ±
Base of log is never negative,
sin θ = 
Therefore, it has only one solution, that means it has unique solution.
The correct answer is a unique solution.

In this question, we have to find type of solution, here, we know if  then we can take antilog, b = c^a, and cos² θ = 1 – 2 sin² θ . Remember these terms and find the solution easily.