If S=(x ∈R : sin ( x+1 x^2+2x+2 )- sin^-1 ( x x^2+1 )= 4)then _x ∈S( sin ((x^2+x+5) 2 )- cos ((x^2+x+5) )) is equal to_________.

Question

Accepted Answer

Correct answer: 4. (4) Given, sin^-1 ⁡ ( x+1 x^2+2x+2 )- sin^-1 ⁡ ( x x^2+1 )= 4 ⇒ sin^-1 ⁡ ( x+1 x^2+x+2 )= 4+ sin^-1 ⁡ x x^2+1 ⇒ ( x+1 x^2+x+2 )= sin ⁡( 4+ sin^-1 ⁡ x x^2+1) ⇒ ( x+1 x^2+x+2 )= 1 2 1 x^2+1+ 1 2 x x^2+1 ⇒ x+1 x^2+x+2= x+1 2 x^2+1 ⇒ (x+1 )( 2 x^2+1- x^2+x+2)=0 ⇒x=-1 or x^2+x+2= 2 x^2+1 Now solving, x^2+x+2= 2 x^2+1 we get, x^2+x+2=2(x^2+1) ⇒x^2-x=0 ⇒x=0 ,x=1 rejected as x=1 will not satisfy the given equation Hence, S= 0 ,1 Now solving, 0.05em 0.05em( sin ⁡(x^2+x+5) 2- cos ⁡(x^2+x+5) ) =( sin ⁡ 5 2- cos ⁡5 + sin ⁡ 5 2- cos ⁡5 ) =1- (-1 )+1- (-1 )=4

$If S = (x \in R : sin (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}) then \sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π)) is equal to_________.$

Solution

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