If 8 sin 2-10 sin-cos-3 cos 2-0 then -

The correct option is D 8 sin^2 θ + 10 sinθcosθ 3 cos^2 θ = 0 ⇒ (2 sinθ + 3 cosθ)(4 sinθ cosθ) = 0 ⇒ 2 sinθ + 3 cosθ = 0, 4 sinθ = cosθ = 0 ⇒tanθ = 3/2, t ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

If 8 sin 2θ+1...

Question

If $8 {sin}^{2} θ + 10 sin θ cos θ - 3 {cos}^{2} θ = 0$ then $θ =$

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Solution

The correct option is D
8 ${sin}^{2} θ + 10 sin θ cos θ - 3 {cos}^{2} θ = 0 \Rightarrow (2 sin θ + 3 cos θ) (4 sin θ - cos θ) = 0$
$\Rightarrow 2 sin θ + 3 cos θ = 0, 4 sin θ = cos θ = 0 \Rightarrow tan θ = \frac{- 3}{2}, tan θ = \frac{1}{4}$
$\Rightarrow θ = n π - T a n^{- 1} (\frac{3}{2}) o r n π + T a n^{- 1} (\frac{1}{4}), n ϵ Z$

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