If x-0- sin-12-x-cos-12-x-

The correct option is C No Solution sin^ 1x+cos^ 1x=π2, x∈[ 1,1] but 2π+x∉[ 1,1] as (2π+x) gt;1 ∴sin^ 1(2π+x)+cos^ 1(2π+x) #160; has ...

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Modulus Function

If x>0, sin...

Question

If $x > 0, {sin}^{- 1} (2 π + x) + {cos}^{- 1} (2 π + x) =$

2 π + \frac{π}{2}

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\frac{π}{2}

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\frac{- π}{2}

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No Solution

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f (x) = {\begin{matrix} x s i n x, & w h e n 0 < x \leq \frac{π}{2} \frac{π}{2} s i n (π + x), & w h e n \frac{π}{2} < x < π \end{matrix}, t h e n

s i n^{- 1} [x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . .] = π / 2 - c o s^{- 1} [x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . .]

0 < | x | < \sqrt{2}

t a n^{- 1} \sqrt{x (x + 1)} + s i n^{- 1} \sqrt{x^{2} + x + 1} = π / 2

- \sqrt{2} < x < 0

{sin}^{- 1} (\frac{2 π}{4})

{cos}^{- 1} (cos \frac{7 π}{6})

{tan}^{- 1} (tan \frac{2 π}{3})

cos ({cos}^{- 1} (\frac{\sqrt{3}}{2}) + \frac{π}{6})

A

0 < x < \frac{π}{2}

{sin}^{- 1} (c o s x) + {cos}^{- 1} (s i n x) = π - 2 x

{cos}^{- 1} x = \frac{π}{2} - {sin}^{- 1} x \forall x \in [0, 1]

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x + a^{2} \sqrt{2} s i n x, & 0 \leq x < \frac{π}{4} x c o t x + b, & \frac{π}{4} \leq x < \frac{π}{2} b s i n 2 x - a c o s 2 x, & \frac{π}{2} \leq x \leq π \end{matrix}

[0, π]

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