Koличecтвo кopнeй уpaвнeния [tex]sin^{2}x-sinx-cos^{2}x=0[/tex], пpинaдлeжaщих oтpeзкy [-3;3], paвнo. Created by benlinus130. algebra-ru

tex], пpинaдлeжaщих oтpeзкy [-3;3], paвнo...

Ilyazov45

$Sin^{2}x-Sinx-Cos^{2}x=0 \\\\Sin^{2}x-Sinx-(1-Sin^{2}x)=0\\\\Sin^{2}x-Sinx-1+Sin^{2}x=0\\\\2Sin^{2}x-Sinx-1=0\\\\Sinx=m \ ; \ -1\leq m\leq 1 \\\\2m^{2} -m-1=0\\\\D=(-1)^{2}-4\cdot 2\cdot(-1)=1+8=9=3^{2} \\\\m_{1}=\dfrac{1-3}{4}=-\dfrac{1}{2}\\\\m_{2}=\dfrac{1+3}{4}=1$

$1) \ Sinx=-\dfrac{1}{2} \\\\\left[\begin{array}{ccc}x=-\dfrac{\pi }{6}+2\pi n \\x=-\dfrac{5\pi }{6}+2\pi n \end{array}\right\\\\x_{1}=-\dfrac{\pi }{6} \ ; \ x_{2}=-\dfrac{5\pi }{6} \\\\2) \ Sinx=1\\\\x=\dfrac{\pi }{2} +2\pi n,n\in Z\\\\x_{3}=\dfrac{\pi }{2}$

Ответ : 3 корня

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Universalka

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Ответ:

равно 3

Объяснение:

На фото два способа отбора корней.

1) Через двойное неравенство

2) Через тригонометрический круг.

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