What is the answer to 1/(sqrt(1-x^2))= pi/2 ?

The answer to 1/(sqrt(1-x^2))= pi/2 is x=(sqrt(-4+pi^2))/pi ,x=-(sqrt(-4+pi^2))/pi

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1/(sqrt(1-x^2))= pi/2

\frac{1}{1 - x ^{2}} = \frac{π}{2}

x = \frac{- 4 + π ^{2}}{π}, x = - \frac{- 4 + π ^{2}}{π}

Decimal Notation

x = 0.77117 \dots, x = - 0.77117 \dots

Solution steps

\frac{1}{1 - x ^{2}} = \frac{π}{2}

Apply fraction cross multiply: if

\frac{a}{b} = \frac{c}{d}

then

a \cdot d = b \cdot c

1 \cdot 2 = 1 - x^{2} π

Switch sides

1 - x^{2} π = 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

1 - x^{2} π = 2

Divide both sides by

π

1 - x^{2} = \frac{2}{π}

Square both sides:

1 - x^{2} = \frac{4}{π ^{2}}

1 - x^{2} = \frac{4}{π ^{2}}

Solve

1 - x^{2} = \frac{4}{π ^{2}} : x = \frac{- 4 + π ^{2}}{π}, x = - \frac{- 4 + π ^{2}}{π}

x = \frac{- 4 + π ^{2}}{π}, x = - \frac{- 4 + π ^{2}}{π}

Verify Solutions:

x = \frac{- 4 + π ^{2}}{π}

True

, x = - \frac{- 4 + π ^{2}}{π}

True

The solutions are

x = \frac{- 4 + π ^{2}}{π}, x = - \frac{- 4 + π ^{2}}{π}

x - 11 = 5

g, t = 2 π \frac{l}{g}

(2 x + 3)^{\frac{3}{4}} - 3 = 5

4 x - 12 + 11 = 19

2 + 2 x - 5 = x

What is the answer to 1/(sqrt(1-x^2))= pi/2 ?
The answer to 1/(sqrt(1-x^2))= pi/2 is x=(sqrt(-4+pi^2))/pi ,x=-(sqrt(-4+pi^2))/pi