LetΔ$$($$θ$$)=1sin$$⁡θ$$1$$−$$sin$$⁡θ$$1sin$$⁡θ−$$1$$−$$sin$$⁡θ$$1$$,$$0$$≤θ≤$$2$$$$π$$Solution of ∆$$($$θ$$)$$ $$=$$ $$3$$ is

Correct option is 2π4,3π4,5π4,7π4

Questions

Let
$Δ (θ) = |\begin{array}{ccc} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{array}|, 0 \leq θ \leq 2 π$
Solution of $∆ (θ) = 3$ is

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detailed solution

Correct option is B

Using C1→C1+C3 we getΔ(θ)=2sin⁡θ101sin⁡θ0−sin⁡θ1=21+sin2⁡θ∴ Δ(θ)=3⇒2sin2⁡θ=1⇒ sin⁡θ=±12⇒θ=π4,3π4,5π4,7π4

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LetΔ(θ)=1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1,0≤θ≤2πSolution of ∆(θ) = 3 is

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Let
$Δ (θ) = |\begin{array}{ccc} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{array}|, 0 \leq θ \leq 2 π$
Solution of $∆ (θ) = 3$ is