How do you solve $2 cos - 1 x = π$ $2 cos-1 x = pi$ ?

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Answer 1 · 2018-02-24T18:48:12

$x = 0$ $color(blue)(x=0)$

I'm reading this as $2 {cos}^{- 1} (x) = π$ $2cos^-1(x)=pi$

$2 {cos}^{- 1} (x) = π$ $2cos^-1(x)=pi$

Divide both sides by 2:

${cos}^{- 1} (x) = \frac{π}{2}$ $cos^-1(x)=pi/2$

Take the cosine of both sides: *

$cos ({cos}^{- 1} (x)) = cos (\frac{π}{2})$ $cos(cos^-1(x))=cos(pi/2)$

$x = cos (\frac{π}{2})$ $x=cos(pi/2)$

$cos (\frac{π}{2}) = 0$ $cos(pi/2)=0$

Hence:

$x = 0$ $color(blue)(x=0)$

*
If $y = cos (x)$ $y=cos(x)$

Then:

$x = {cos}^{- 1} (y)$ $x=cos^-1(y)$

$cos ({cos}^{- 1} (y)) = cos (x) = y$ $cos(cos^-1(y))=cos(x)=y$

How do you solve 2 cos-1 x = pi2cos−1x=π2 cos-1 x = pi?