How do you solve $\frac{1}{cos x - sin x} = cos x + sin x$ $1/(cosx-sinx)=cosx+sinx$ ?

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How do you solve $\frac{1}{cos x - sin x} = cos x + sin x$ $1/(cosx-sinx)=cosx+sinx$ ?

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Answer 1 · 2016-12-07T04:50:39

$x = 0 + 2 π n, π + 2 π n, cos x - sin x \neq 0$ $x = 0 + 2pin, pi + 2pin, cosx - sinx !=0$

Explanation:

Make the equation into a product.

$1 = (cos x + sin x) (cos x - sin x)$ $1 = (cosx + sinx)(cosx - sinx)$

$1 = {cos}^{2} x - {sin}^{2} x$ $1 = cos^2x - sin^2x$

Solve the pythagorean identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$ for ${cos}^{2} x$ $cos^2x$ and substitute.

$1 = 1 - {sin}^{2} x - {sin}^{2} x$ $1 = 1- sin^2x - sin^2x$

$0 = - 2 {sin}^{2} x$ $0 = -2sin^2x$

$0 = sin x$ $0 = sinx$

$x = 0 + 2 π n, π + 2 π n$ $x = 0 + 2pin, pi + 2pin$

Our restrictions are $cos x - sin x \neq 0$ $cosx- sinx != 0$ .

Hopefully this helps!

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How do you solve $\frac{1}{cos x - sin x} = cos x + sin x$ $1/(cosx-sinx)=cosx+sinx$ ?

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How do you solve $\frac{1}{cos x - sin x} = cos x + sin x$ $1/(cosx-sinx)=cosx+sinx$ ?