How do you solve $1/(cosx-sinx)=cosx+sinx$ ?

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Answer 1 · 2016-12-07T04:50:39

$x = 0 + 2pin, pi + 2pin, cosx - sinx !=0$

Make the equation into a product.

$1 = (cosx + sinx)(cosx - sinx)$

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

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

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

$0 = -2sin^2x$

$0 = sinx$

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

Our restrictions are $cosx- sinx != 0$ .

Hopefully this helps!

How do you solve 1/(cosx-sinx)=cosx+sinx1/(cosx-sinx)=cosx+sinx?