Respuesta :

dalendrk
[tex]\sin^2x=\cos^2x\ \ \ \ |we\ know:\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\sin^2x=1-\sin^2x\ \ \ \ \ |add\ \sin^2x\ to\ both\ sides\\\\2\sin^2x=1\ \ \ \ |divide\ both\ sides\ by\ 2\\\\\sin^2x=\dfrac{1}{2}\iff\sin x=-\sqrt{\dfrac{1}{2}}\ or\ \sin x=\sqrt{\dfrac{1}{2}}\\\\x=-\dfrac{\sqrt2}{2}\ or\ x=\dfrac{\sqrt2}{2}\\\\Answer:\ \boxed{x=\frac{\pi}{4}\ or\ x=\frac{3\pi}{4}\ or\ x=\frac{5\pi}{4}\ or\ x=\frac{7\pi}{4}}[/tex]

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