а) 4sin^2 x * cosx+2√3sin2x+3cosx=0б) Найдите корни принадлежащие отрезку [-4п, -2,5п]. Created by khantimirsila001001. algebra-ru

а) 4sin^2 x * cosx+2√3sin2x+3cosx=0б) Найдите корни принадлежащие отрезку [-4п, ...

khantimirsila001001 @khantimirsila001001

July 2022 1 0 Report

а) 4sin^2 x * cosx+2√3sin2x+3cosx=0
б) Найдите корни принадлежащие отрезку [-4п, -2,5п]

Answers & Comments

Universalka

Verified answer

$4Sin^{2}x\cdot Cosx+2\sqrt{3}Sin2x+3Cosx=0\\\\4Sin^{2}x\cdot Cosx+2\sqrt{3}\cdot 2Sinx Cosx+3Cosx=0\\\\4Sin^{2}x\cdot Cosx+4\sqrt{3}Sinx Cosx+3Cosx=0\\\\Cosx(4Sin^{2}x+4\sqrt{3}Sinx+3)=0\\\\Cosx(2Sinx+\sqrt{3})^{2}=0\\\\\left[\begin{array}{ccc}Cosx=0\\2Sinx+\sqrt{3}=0 \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\frac{\pi }{2}+\pi n,n\in Z \\Sinx=-\dfrac{\sqrt{3} }{2} \end{array}\right$

$Sinx=-\frac{\sqrt{3} }{2}\\\\\left[\begin{array}{ccc}x=-\dfrac{\pi }{3}+2\pi n,n\in Z\\x=\dfrac{4\pi }{3} +2\pi n,n\in Z \end{array}\right\\\\1)x=\dfrac{\pi }{2}+\pi n \\\\-4\pi\leq \dfrac{\pi }{2}+\pi n\leq -2,5\pi|\cdot\dfrac{2}{\pi } \\\\-8\leq 1+2n\leq-5\\\\-9\leq 2n\leq-6\\\\-4,5\leq n\leq-3\\\\n=-4 \ \Rightarrow \ \dfrac{\pi }{2}-4\pi =\boxed{-\frac{7\pi }{2}} \\\\n=-3 \ \Rightarrow \ \dfrac{\pi }{2}-3\pi=\boxed{-\dfrac{5\pi }{2}} \\\\\\2)x=-\dfrac{\pi }{3}+2\pi n$

$-4\pi\leq -\dfrac{\pi }{3}+2\pi n\leq -2,5\pi \ |\cdot\dfrac{3}{\pi } \\\\-12\leq -1+6n\leq-7,5\\\\-11\leq 6n\leq -6,5\\\\-1\frac{5}{6} \leq n\leq-1\frac{1}{12} \\\\kornei \ net \\\\\\3)x=\frac{4\pi }{3} +2\pi n\\\\-4\pi \leq \frac{4\pi }{3} +2\pi n \leq -2,5\pi \ |\cdot\dfrac{3}{\pi }\\\\-12\leq 4+6n\leq-7,5\\\\-16\leq6n\leq-13,5\\\\-2\frac{2}{3} \leqn\leq -2,25\\\\kornei \ net\\\\Otvet:\boxed{-\dfrac{7\pi }{2} \ ; \ -\dfrac{5\pi }{2}}$

0 votes Thanks 0

Answers & Comments

Verified answer

рекомендуемые вопросы

Helpful Links

Helpful Social