 # How do you solve \sin^2x-\cos^2x=0 for x in the interval kerrum75 2021-12-13 Answered
How do you solve ${\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x=0$ for x in the interval $\left[0,2\pi \right)$ ?
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Applying the identity ${\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)=\mathrm{cos}\left(2x\right)$ we have
${\mathrm{sin}}^{2}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)=-\mathrm{cos}\left(2x\right)$
In general,
$\mathrm{cos}\left(u\right)=0\to u=\frac{n\pi }{2}$ for some $n\in \mathbb{Z}$
Thus we have
${\mathrm{sin}}^{2}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)=0$
$\to 2x=\frac{n\pi }{2}$ for $n\in \mathbb{Z}$
$⇒x=\frac{n\pi }{4}$ for $n\in \mathbb{Z}$
Restricting our values to the interval $\left[0,2\pi \right]$ gives our final result:
$x\in \left(\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\right)$
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