Question

# Consider the system of linear equations in x, y and z ; (sin 3θ) x - y + z = 0 ......(i) (cos 2θ)x + 4y + 3z = 0 ......(ii) 2x + 7y+ 7z = 0 ......(iii) The value of θ for which the system has nontrivial solution is

A

θ = nπ, nπ + (-1)n (π/6), where n ϵ I.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

θ= nπ+(π/6),, nπ + (-1)n (π/4), where n ϵ I.

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

θ= nπ-π/6, nπ + (-1)n (π/3), where n ϵ I.

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

θ=2nπ , where n ϵ I.

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A θ = nπ, nπ + (-1)n (π/6), where n ϵ I. Eliminating x, y, z from the system of linear equations(i) , (ii), (iii), then ∣∣ ∣∣sin3θ−11cos2θ43277∣∣ ∣∣ =0 or(28−21)sin3θ−(−7−7)cos2θ+2(−3−4)=0or,7sin3θ+14cos2θ−14=0or,sin3θ+cos2θ−2=0or,(3sinθ−4sin3θ)+2(1−2sin2θ)−2=0or4sin3θ+4sin2θ−3sinθ=0orsinθ(4sin2θ+4sinθ−3)=0orsinθ(2sinθ−1)(2sinθ+3)=0 Either sin θ = 0 or sin θ =12 or sinθ=−32 but sinθ = 0, or sinθ=12 is possible or sinθ=−32 is not possible Now, sinθ = 0 or, θ = nπ where n ∈ I. And sin θ =12 = sin(π6), or, θ = nπ +(−1)nπ6, where n ∈ I. Hence the required values of π are θ=nπ, nπ + (−1)n(π6), where n ∈ I.

Suggest Corrections
0
Join BYJU'S Learning Program
Select...
Related Videos
Solving Trigonometric Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
Select...