# Solve the equation on the interval [0,2pi ) sin^2 x-cos^2 x=0

Solve the equation on the interval $\left[0,2\pi \right)$
${\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x=0$
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bardalhg
$\left(1-{\mathrm{cos}}^{2}x\right)-{\mathrm{cos}}^{2}x=0$
$1-2{\mathrm{cos}}^{2}x=0$
$-2{\mathrm{cos}}^{2}x=-1$
${\mathrm{cos}}^{2}x=-\frac{1}{2}$
$\mathrm{cos}x=±\sqrt{\frac{1}{2}}=±\frac{\sqrt{2}}{2}$
$x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$
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stylaria3y
${\mathrm{sin}}^{2}x-\left(1-{\mathrm{sin}}^{2}x\right)=0$, use trig identity
$2{\mathrm{sin}}^{2}x=1$

$x=\frac{\pi }{4}+\frac{n\pi }{2}$
and $x=\frac{3\pi }{4}+\frac{n\pi }{2}$