JEE Main 10 Jan 2019 Shift 2 Solved Paper

|

1	3	7
−1	4	7
sin3θ	cos2θ	2

| = 0
(8 – 7 cos 2θ) – 3(–2 – 7 sin 3θ) + 7 (– cos 2θ – 4 sin 3θ) = 0
14 – 7 cos 2θ + 21 sin 3θ – 7 cos 2θ– 28 sin 3θ = 0
14 – 7 sin 3θ – 14 cos 2θ = 0
14 – 7 (3 sin θ – 4 sin3 θ) – 14 (1 – 2 sin2 θ) = 0
–21 sin θ + 28 sin3 θ + 28 sin2 θ = 0
7 sin θ [–3 + 4 sin2 θ + 4 sin θ] = 0
sin θ,
4 sin2 θ + 6 sin θ – 2 sin θ – 3 = 0
2 sin θ(2 sin θ + 3) – 1 (2 sin θ + 3) = 0
sin θ = -

3

2

; sin θ =

1

2

Hence, 2 solutions in (0, π)