How do you solve \sin^2x-\cos^2x=0 for x in the interval

kerrum75 2021-12-13 Answered

How do you solve

\sin^{2} x - \cos^{2} x = 0

for x in the interval

[0, 2 π)

?

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Expert Answer

usumbiix

Answered 2021-12-14 Author has 33 answers

Applying the identity

\cos^{2} (x) - \sin^{2} (x) = \cos (2 x)

we have

\sin^{2} (x) - \cos^{2} (x) = - \cos (2 x)

In general,

\cos (u) = 0 \to u = \frac{n π}{2}

for some

n \in Z

Thus we have

\sin^{2} (x) - \cos^{2} (x) = 0

\to 2 x = \frac{n π}{2}

for

n \in Z

\Rightarrow x = \frac{n π}{4}

for

n \in Z

Restricting our values to the interval

[0, 2 π]

gives our final result:

x \in (\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4})

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lovagwb

Answered 2021-12-15 Author has 50 answers

Yes! Of course!

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Elementary geometry

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