The number of θ ∈(0,4 π) for which the system of linear equations
3(sin 3 θ) x-y+z=2 ; ; 3(cos 2 θ) x+4 y+3 z=3 ; ; 6 x+7 y+7 z=9
has no solution, is :

Solution

Maths sir · 5 steps

No solution needs two things: D = 0, and then the equations must contradict each other.

1. Set the determinant. Expand along the first row: 3 sin 3θ(28-21) + (21 cos 2θ - 18) + (21 cos 2θ - 24). For the system to have no unique solution this must be zero.

   D = 21 sin 3θ + 42 cos 2θ - 42 = 0

2. Reduce to sin θ. Divide by 21 and use sin 3θ = 3s - 4s³ and 2 cos 2θ - 2 = -4s², with s = sin θ. This gives s(4s² + 4s - 3) = 0, that is s(2s-1)(2s+3) = 0, so sin θ = 0 or sin θ = 1/2.
3. Check for contradiction. If sin θ = 0: equation 3 minus twice equation 2 gives -y + z = 3, but equation 1 says -y + z = 2. If sin θ = 1/2: sin 3θ = 1, cos 2θ = 1/2, and equation 1 plus twice equation 2 has the left side 6x + 7y + 7z but the right side 8, not 9. Both cases are inconsistent.
4. Count in the interval. sin θ = 0 in (0, 4π): π, 2π, 3π, which is 3 values (0 and 4π are excluded). sin θ = 1/2: π/6, 5π/6, 13π/6, 17π/6, which is 4 values.
5. So. The system has no solution for 3 + 4 = 7 values of θ, which is option (b).

The idea to keep. For no solution, D = 0 is only the first filter. Then find two equations whose combination matches the third on the left side but gives a different right side.

Answer from the NTA final answer key. Solution by PrepWiser.