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Question Number 203625 by professorleiciano last updated on 23/Jan/24

	Answered by esmaeil last updated on 23/Jan/24
	$1+(((sin^2 x+cos^2 x)^2 −2sin^2 xcos^2 x)/(sin^2 xcos^2 x))=3→ 1=sin^2 2x→sin2x=±1=sin(±(π/2)) 2x=kπ±(π/2)→ { ((2x=2kπ+(π/2))),((2x=2kπ+π−(π/2))) :} { ((2x=2kπ−(π/2))),((2x=2kπ+((3π)/2))) :} x=kπ±(π/4) ,x=kπ+((3π)/4)$
	$1 + \frac{{({s i n}^{2} x + {c o s}^{2} x)}^{2} - 2 {s i n}^{2} {x c o s}^{2} x}{{s i n}^{2} {x c o s}^{2} x} = 3 \to$ $1 = {s i n}^{2} 2 x \to s i n 2 x = \pm 1 = s i n (\pm \frac{π}{2})$ $2 x = k π \pm \frac{π}{2} \to {\begin{cases} 2 x = 2 k π + \frac{π}{2} \\ 2 x = 2 k π + π - \frac{π}{2} \end{cases}$ ${\begin{cases} 2 x = 2 k π - \frac{π}{2} \\ 2 x = 2 k π + \frac{3 π}{2} \end{cases}$ $x = k π \pm \frac{π}{4}, x = k π + \frac{3 π}{4}$
	Commented by professorleiciano last updated on 24/Jan/24
	Sua resposta está totalmente errada.
	Answered by professorleiciano last updated on 23/Jan/24

	Commented by esmaeil last updated on 23/Jan/24

	${(s e c x)}^{2} \overset{!!!!! (w h y)}{=} {s e c x}^{2}$
	Commented by esmaeil last updated on 24/Jan/24

	$w h y d o y o u o n l y a n s w e r y o u r o w n$ $q u e s t i o n . y o u r a n s w e r i s a l s o w r o n g .$
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