Exams › JEE Main › Maths
For the variables x, y and z, consider the homogeneous linear system
x sin 3θ − y + z = 0,
x cos 2θ + 4y + 3z = 0,
2x + 7y + 7z = 0.
If a non-zero solution exists, then for an integer n, the possible values of θ are
- π(n + (−1)ⁿ/3)
- π(n + (−1)ⁿ/4)
- π(n + (−1)ⁿ/6)
- nπ/2
Correct answer: π(n + (−1)ⁿ/6)
Solution
For a non-trivial solution the determinant must vanish: -28 sin^2(theta) + 7 sin(3theta) = 0. Using sin(3theta)=3 sin-4 sin^3 gives 7 sin(theta)(3 - 4 sin^2(theta) - 4 sin(theta)) = 0, so sin(theta)=1/2 (the other root -3/2 is invalid). Thus theta = n*pi + (-1)^n * pi/6 = pi(n + (-1)^n/6).
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