If 2cos-1-1-x2-2- then x-

2cos^ 1√(1+x2)=π2 ⇒cos^ 1√(1+x2)=π4 ⇒1+x2=12 ∴ x=0 ...

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If 2cos-1√1+x...

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If $2 {cos}^{- 1} \sqrt{\frac{1 + x}{2}} = \frac{π}{2}$ , then $x =$

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{tan}^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}) = \frac{π}{4} - \frac{1}{2} {cos}^{- 1} x, - \frac{1}{\sqrt{2}} \leq x \leq 1

p u t x = cos 2 θ]

2 c o s^{- 1} \sqrt{\frac{1 + x}{2}} = \frac{π}{2}

0 \leq {cos}^{- 1} x \leq π

- \frac{π}{2} \leq {sin}^{- 1} x \leq \frac{π}{2}

cos ({sin}^{- 1} x + 2 {cos}^{- 1} x)

x = \frac{1}{5}

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{s i n 2 x}{\sqrt{1 - c o s 2 x}} f o r 0 < x \leq π / 2 \frac{c o s}{π - 2 x} f o r \frac{π}{2} < x < π \end{matrix}

x = π / 2

x = t a n [\frac{π}{4} + \frac{1}{2} c o s^{- 1} \frac{a}{b}]

y = t a n [\frac{π}{4} - \frac{1}{2} c o s^{- 1} \frac{a}{b}]

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