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Question: The values of \(\theta ,\lambda \) for which the following equations, \(\begin{aligned} & \si...

The values of θ,λ\theta ,\lambda for which the following equations,
sin⁡θx−cos⁡θy+(λ+1)z=0 cos⁡θx+sin⁡θy−λz=0 λx+(λ+1)y+cos⁡z=0 \begin{aligned} & \sin \theta x-\cos \theta y+\left( \lambda +1 \right)z=0 \\\ & \cos \theta x+\sin \theta y-\lambda z=0 \\\ & \lambda x+\left( \lambda +1 \right)y+\cos z=0 \\\ \end{aligned}
Have non trivial solution is,
A. \theta =n\pi ,\lambda \in R-\left\\{ 0 \right\\}
B. θ=2nπ,λ\theta =2n\pi ,\lambda is any rational number
C. θ=(2n+1)π,λ∈R,n∈I\theta =\left( 2n+1 \right)\pi ,\lambda \in R,n\in I
D. 2(n+1)π2,λ∈R,n∈I2\left( n+1 \right)\dfrac{\pi }{2},\lambda \in R,n\in I

Explanation

Solution

Hint: To solve the above question, we will make use of the concept that if the system of equations in which the determinant of the coefficient is equal to zero, then the system of equations has a non-trivial solution. After forming the determinant, we can expand it, equate it to zero and then we will get the value for cos⁡θ\cos \theta . Then we can use the general formula for cos⁡θ=cos⁡α\cos \theta =\cos \alpha as θ=2nπ±α\theta =2n\pi \pm \alpha .

Complete step-by-step answer:
We have been asked to find the values of θ,λ\theta ,\lambda for which the following equations, have non-trivial solution,
sin⁡θx−cos⁡θy+(λ+1)z=0 cos⁡θx+sin⁡θy−λz=0 λx+(λ+1)y+cos⁡z=0 \begin{aligned} & \sin \theta x-\cos \theta y+\left( \lambda +1 \right)z=0 \\\ & \cos \theta x+\sin \theta y-\lambda z=0 \\\ & \lambda x+\left( \lambda +1 \right)y+\cos z=0 \\\ \end{aligned}
Since, we know that the system of equation in which the determinant of the coefficient is equal to zero, then the system of equation has non-trivial solution. So, we will form a determinant and enter all the coefficients of x, y and the constant term as shown below.
∣sin⁡θ −cos⁡θ λ+1 cos⁡θ sin⁡θ −λ  λ λ+1 cos⁡θ ∣=0\left| \begin{aligned} & \sin \theta \text{ }-\cos \theta \text{ }\lambda +1 \\\ & \cos \theta \text{ }\sin \theta \text{ }-\lambda \\\ & \text{ }\lambda \text{ }\lambda +1\text{ }\cos \theta \\\ \end{aligned} \right|=0
Now, on expanding the determinants, we get the terms as,
sin⁡θ[sin⁡θcos⁡θ−(−λ)(λ+1)]+cos⁡θ(cos⁡2θ+λ2)+(λ+1)[(λ+1)cos⁡θ−λsin⁡θ]=0 ⇒sin⁡2θcos⁡θ+λ(λ+1)sin⁡θ+cos⁡3θ+λ2cos⁡θ+(λ+1)2cos⁡θ−λ(λ+1)sin⁡θ=0 \begin{aligned} & \sin \theta \left[ \sin \theta \cos \theta -\left( -\lambda \right)\left( \lambda +1 \right) \right]+\cos \theta \left( {{\cos }^{2}}\theta +{{\lambda }^{2}} \right)+\left( \lambda +1 \right)\left[ \left( \lambda +1 \right)\cos \theta -\lambda \sin \theta \right]=0 \\\ & \Rightarrow {{\sin }^{2}}\theta \cos \theta +\lambda \left( \lambda +1 \right)\sin \theta +{{\cos }^{3}}\theta +{{\lambda }^{2}}\cos \theta +{{\left( \lambda +1 \right)}^{2}}\cos \theta -\lambda \left( \lambda +1 \right)\sin \theta =0 \\\ \end{aligned}
We will substitute sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta in the above expression. So, we get,
(1−cos⁡2θ)cos⁡θ+λ(λ+1)sin⁡θ+cos⁡3θ+λ2cos⁡θ+(λ+1)2cos⁡θ−λ(λ+1)sin⁡θ=0\left( 1-{{\cos }^{2}}\theta \right)\cos \theta +\lambda \left( \lambda +1 \right)\sin \theta +{{\cos }^{3}}\theta +{{\lambda }^{2}}\cos \theta +{{\left( \lambda +1 \right)}^{2}}\cos \theta -\lambda \left( \lambda +1 \right)\sin \theta =0
By rearranging and cancelling the similar terms, we get,
(1−cos⁡2θ)cos⁡θ+cos⁡3θ+λ2cos⁡θ+(λ+1)2cos⁡θ=0\left( 1-{{\cos }^{2}}\theta \right)\cos \theta +{{\cos }^{3}}\theta +{{\lambda }^{2}}\cos \theta +{{\left( \lambda +1 \right)}^{2}}\cos \theta =0
Now, we will take cos⁡θ\cos \theta common form the above expression. So, we will get,
cos⁡θ[1−cos⁡2θ+cos⁡2θ+λ2+(λ+1)2]=0 ⇒cos⁡θ[1+λ2+(λ+1)2]=0 \begin{aligned} & \cos \theta \left[ 1-{{\cos }^{2}}\theta +{{\cos }^{2}}\theta +{{\lambda }^{2}}+{{\left( \lambda +1 \right)}^{2}} \right]=0 \\\ & \Rightarrow \cos \theta \left[ 1+{{\lambda }^{2}}+{{\left( \lambda +1 \right)}^{2}} \right]=0 \\\ \end{aligned}
Since, 1+λ2+(λ+1)21+{{\lambda }^{2}}+{{\left( \lambda +1 \right)}^{2}} cannot be equal to zero, that means, cos⁡θ=0\cos \theta =0. We know that the general solution for cos⁡θ=cos⁡α\cos \theta =\cos \alpha is given by,
θ=2nπ±α ⇒cos⁡θ=cos⁡π2 ⇒θ=2nπ±π2 or θ=(2n+1)π2 \begin{aligned} & \theta =2n\pi \pm \alpha \\\ & \Rightarrow \cos \theta =\cos \dfrac{\pi }{2} \\\ & \Rightarrow \theta =2n\pi \pm \dfrac{\pi }{2} \\\ & or\text{ }\theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\\ \end{aligned}
Hence, λ\lambda can take any values, so, λ∈R\lambda \in R and θ=(2n+1)π2\theta =\left( 2n+1 \right)\dfrac{\pi }{2}, where n∈In\in I (integer).
Therefore, the correct option is option D.

Note: The students must be careful while doing the calculations, as there are a lot of calculations involved in this question and should be careful with the signs too in each step. The students might stop after finding cos⁡θ=0\cos \theta =0 and may directly write the answer as θ=90\theta =90. But this is not correct, they must think of the general solution as there can be many values for θ\theta . Also, they should mention that λ\lambda is an element of R.