Q. 52 I5.0( 2 Votes )

Solve the followi

Answer :


Let


We need to find the roots of Δ = 0.


Recall that the value of a determinant remains same if we apply the operation Ri Ri + kRj or Ci Ci + kCj.


Applying C1 C1 + C2, we get




Applying R2 R2 – R1, we get




Applying R3 R3 – 3R1, we get




Expanding the determinant along C1, we have


Δ = (1)[(10)(2 – 3sin(3θ)) – (20)(cos(2θ) – sin(3θ))]


Δ = [20 – 30sin(3θ) – 20cos(2θ) + 20sin(3θ)]


Δ = 20 – 10sin(3θ) – 20cos(2θ)


From trigonometry, we have sin(3θ) = 3sinθ – 4sin3θ and cos(2θ) = 1 – 2sin2θ.


Δ = 20 – 10(3sinθ – 4sin3θ) – 20(1 – 2sin2θ)


Δ = 20 – 30sinθ + 40sin3θ – 20 + 40sin2θ


Δ = –30sinθ + 40sin2θ + 40sin3θ


Δ = 10(sinθ)(–3 + 4sinθ + 4sin2θ)


The given equation is Δ = 0.


10(sinθ)(–3 + 4sinθ + 4sin2θ) = 0


(sinθ)(–3 + 4sinθ + 4sin2θ) = 0


Case – I:


sin θ = 0 θ = kπ, where k ϵ Z


Case – II:


–3 + 4sinθ + 4sin2θ = 0


4sin2θ + 4sinθ – 3 = 0


4sin2θ + 6sinθ – 2sinθ – 3 = 0


2sinθ(2sinθ + 3) – 1(2sinθ + 3) = 0


(2sinθ – 1)(2sinθ + 3) = 0


2sinθ – 1 = 0 or 2sinθ + 3 = 0


2sinθ = 1 or 2sinθ = –3


or


However, as –1 ≤ sin θ ≤ 1.



, where k ϵ Z


Thus, kπ and for all integral values of k are the roots of the given determinant equation.


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