Let f(θ)=|- sin θ 1 0 1 0 sin θ sin θ 1 1|. Number of solutions of the equation f(θ)=0 in (0,2 π) is

Question

Let f(θ)=|- sin θ 1 0 1 0 sin θ sin θ 1 1|. Number of solutions of the equation f(θ)=0 in (0,2 π) is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Clearly, f(θ)=2 sin 2 θ-1=- cos 2 θ ∴ f (θ)=0 ⇒ cos 2 θ=0 ⇒ θ=(π/4), (3 π/4), (5 π/4), (7 π/4) So, number of solution are 4.

Q. Let $f (θ) = ∣ ∣ - sin θ 1 sin θ 101 0 sin θ 1 ∣ ∣$ . Number of solutions of the equation $f (θ) = 0$ in $(0, 2 π)$ is