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Find the value of θ satisfying [11sin3θ-43cos2θ7-7-2] = 0 - Mathematics

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Question

Find the value of θ satisfying `[(1, 1, sin3theta),(-4, 3, cos2theta),(7, -7, -2)]` = 0

Sum
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Solution

We have, `[(1, 1, sin3theta),(-4, 3, cos2theta),(7, -7, -2)]` = 0

Expanding along C3, we get

`sin 3theta xx (28 - 21) - cos 2theta xx (7 - 7) - 2(3 + 4)` = 0

⇒ `7 sin 3theta + 14 cos 2theta - 14` = 0

⇒ `sin 3theta + 2 cos 2theta - 2` = 0

⇒ `(3 sin theta - 4 sin^3 theta) + 2(1 - 2 sin^2 theta) - 2` = 0

⇒ `4 sin^3 theta - 4 sin^2 theta + 3 sin theta` = 0

⇒ `sin theta(4 sin^2 theta - 4 sin theta + 3)` = 0

⇒ `sin theta(4 sin^2 theta - 6 sin theta + 2 sin theta + 3)` = 0

⇒ `sin theta (2 sin theta + 1)(2 sin theta - 3)` = 0

⇒ sin θ or sin θ = `(-1)/2` or sin θ = `3/2`

⇒ θ = `"n"pi` or θ = `"m"pi + (-1)^"n" (-pi/6); "m", "n" ∈ "Z"`

sin θ = `(-3)/2` is not possible.

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Chapter 4: Determinants - Exercise [Page 78]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 4 Determinants
Exercise | Q 12 | Page 78

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