The product of all real values of x satisfying the equationsin−1cos(x2+5∣x∣+32x2+10∣x∣+4)=cot(cot−1(9∣x∣2−18∣x∣))+2π is

sin−1cos(x2+5x+32x2+10x+4 )=cot(cot−1(9x2−18x ) )+2π ⇒sin−1sin(2π −(x2+5x+32x2+10x+4 ) )=9x2−18x+2π ⇒2π −x2+5x+32x2+10x+4=9x2−18x+2π ⇒−18x3−90x2−36x=2x2+10x+6−18x3−90x−54x⇒2x2−8x+6=0⇒x2−4+3=0⇒(x−1)(x−3)=0Therefore product of roots is 1×3=3

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The product of all real values of x satisfying the equation ^{

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The product of all real values of x satisfying the equation
$sin^{- 1} cos (\frac{2 x ^{2} + 10 ∣ x ∣ + 4}{x ^{2} + 5 ∣ x ∣ + 3}) = cot (cot^{- 1} (\frac{2 - 18 ∣ x ∣}{9 ∣ x ∣})) + \frac{π}{2}$ is

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The product of all real values of x satisfying the equation

sin^{- 1} cos (\frac{2 x ^{2} + 10 ∣ x ∣ + 4}{x ^{2} + 5 ∣ x ∣ + 3}) = cot (cot^{- 1} (\frac{2 - 18 ∣ x ∣}{9 ∣ x ∣})) + \frac{π}{2}

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s i n^{- 1} cos (\frac{2 x ^{2} + 10 x + 4}{x ^{2} + 5 x + 3}) = co t (co t^{- 1} (\frac{2 - 18 x}{9 x})) + \frac{π}{2} \Rightarrow s i n^{- 1} s in (\frac{π}{2} - (\frac{2 x ^{2} + 10 x + 4}{x ^{2} + 5 x + 3})) = \frac{2 - 18 x}{9 x} + \frac{π}{2} \Rightarrow \frac{π}{2} - \frac{2 x ^{2} + 10 x + 4}{x ^{2} + 5 x + 3} = \frac{2 - 18 x}{9 x} + \frac{π}{2} \Rightarrow - 18 x^{3} - 90 x^{2} - 36 x = 2 x^{2} + 10 x + 6 - 18 x^{3} - 90 x - 54 x \Rightarrow 2 x^{2} - 8 x + 6 = 0 \Rightarrow x^{2} - 4 + 3 = 0 \Rightarrow (x - 1) (x - 3) = 0

Therefore product of roots is

1 \times 3 = 3

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Question Text	The product of all real values of x satisfying the equation $sin^{- 1} cos (\frac{2 x ^{2} + 10 ∣ x ∣ + 4}{x ^{2} + 5 ∣ x ∣ + 3}) = cot (cot^{- 1} (\frac{2 - 18 ∣ x ∣}{9 ∣ x ∣})) + \frac{π}{2}$ is
Answer Type	Text solution:1
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The product of all real values of x satisfying the equationsin−1cos(x2+5∣x∣+32x2+10∣x∣+4​)=cot(cot−1(9∣x∣2−18∣x∣​))+2π​ is

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The product of all real values of x satisfying the equation
$sin^{- 1} cos (\frac{2 x ^{2} + 10 ∣ x ∣ + 4}{x ^{2} + 5 ∣ x ∣ + 3}) = cot (cot^{- 1} (\frac{2 - 18 ∣ x ∣}{9 ∣ x ∣})) + \frac{π}{2}$ is