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# precalculus,

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Find all values of $$x$$ such that $$0\leq x\leq 2\pi$$ and $$\sin(x)-\cos(x) = \dfrac{1}{\sqrt{2}}$$.

been stuck on this for a while now, help would be appreciated!

Nov 10, 2021

#1
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$\sin x - \cos x = \frac{1}{\sqrt{2}}$

$1 - \sin 2x = \frac{1}{2}$

$\sin 2x = \frac{1}{2}$

$2x = \frac{\pi}{6}$ or $2x = \frac{5 \pi}{6}$

$x = \frac{\pi}{12}$ or $x = \frac{5 \pi}{12}$

Nov 10, 2021
#2
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Thanks for answering guest but you need to put your latex in a latex box for it to render properly.

Nov 12, 2021