The number of solution satisfying the question -left - beginmatrix 1-sin - 3 - 1-cos - 3- theta - 1cos - theta - 4 -cos- theta 0 - 3 -sin- theta

Sin^3 θ + cos^3 θ + 1 = 3 sin θ cos θ ⇒ sin θ + cos θ + 1 = 0 or sin θ = cos θ =1 ⇒ sin θ + cos θ +1=0 ⇒θ =π, 3 π/2, 3 π, 7 π/2 in θϵ (0, 4 π) ...

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Standard XII

Mathematics

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

The number of...

Question

The number of solution satisfying the question $∣ ∣ ∣ \begin{matrix} 1 + s i n^{3} θ & 1 - c o s^{3} θ & 1 \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} c o s θ & 4 c o s θ 0 & 3 s i n θ \end{matrix} ∣ ∣ ∣ i n θ ϵ (0, 4 π)$ is equal to ___

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Solution

$s i n^{3} θ + c o s^{3} θ + 1 = 3 s i n θ c o s θ \Rightarrow s i n θ + c o s θ + 1 = 0 o r s i n θ = c o s θ = 1 \Rightarrow s i n θ + c o s θ + 1 = 0 \Rightarrow θ = π, \frac{3 π}{2}, 3 π, \frac{7 π}{2} i n θ ϵ (0, 4 π)$

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Q. The solution of

\frac{3 sin θ - sin 3 θ}{1 + cos θ} + \frac{3 cos θ + cos 3 θ}{1 - sin θ} = 4 \sqrt{2} cos (θ + \frac{π}{4})

$Solve the following equations : (i) c o s θ + c p s 2 θ + c o s 3 θ = 0 (i i) c o s θ + c o s 3 θ - c o s 2 θ = 0 (i i i) s i n θ + s i n 5 θ = s i n 3 θ (i v) c o s θ c o s 2 θ c o s 3 θ = \frac{1}{4} (v) c o s θ + s i n θ = c o s 2 θ + s i n 2 θ (v i) s i n θ + s i n 2 θ + s i n 3 θ = 0 (v i i) s i n θ + s i n 2 θ + s i n 3 θ + s i n 4 θ = 0 (v i i i) s i n 3 θ - s i n θ = 4 c o s^{2} θ - 2 (i x) s i n 2 θ - s i n 4 θ + s i n 6 θ = 0$

Q. Let

θ \in (0, \frac{π}{2}) .

If the system of linear equations,

\begin{matrix} (1 + {cos}^{2} θ) x + {sin}^{2} θ y + 4 sin 3 θ z = 0 {cos}^{2} θ x + (1 + {sin}^{2} θ) y + 4 sin 3 θ z = 0 {cos}^{2} θ x + {sin}^{2} θ y + (1 + 4 sin 3 θ) z = 0 \end{matrix}

has a non-trivial solution, then the value of

θ

Q. I:

sin θ . sin (60^{0} - θ) . sin (60^{0} + θ) = \frac{1}{4} sin 3 θ

II:

cos θ cos (120^{0} - θ) cos (120^{0} + θ) =

\frac{1}{4} cos 3 θ

2 ({sin}^{6} θ + {cos}^{6} θ) - 3 ({sin}^{4} θ + {cos}^{4} θ) + 1 = 0

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The number of solution satisfying the question ∣∣∣1+sin3θ1−cos3θ1∣∣∣=∣∣∣cos θ4 cos θ03 sin θ∣∣∣ in θ ϵ (0, 4π) is equal to ___

sin3θ+cos3θ+1=3sin θ cos θ⇒sinθ+cosθ+1=0 or sinθ=cosθ=1⇒sinθ+cosθ+1=0⇒θ=π, 3π2, 3π, 7π2 in θϵ(0,4π)

The number of solution satisfying the question $∣ ∣ ∣ \begin{matrix} 1 + s i n^{3} θ & 1 - c o s^{3} θ & 1 \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} c o s θ & 4 c o s θ 0 & 3 s i n θ \end{matrix} ∣ ∣ ∣ i n θ ϵ (0, 4 π)$ is equal to ___

$s i n^{3} θ + c o s^{3} θ + 1 = 3 s i n θ c o s θ \Rightarrow s i n θ + c o s θ + 1 = 0 o r s i n θ = c o s θ = 1 \Rightarrow s i n θ + c o s θ + 1 = 0 \Rightarrow θ = π, \frac{3 π}{2}, 3 π, \frac{7 π}{2} i n θ ϵ (0, 4 π)$