[tex] 5^\frac12 + 5^{\frac12+log_{5}sin{\frac{ \pi x}{3}}} = 15^{\frac12+log_{15}cos{\frac{ \pi x}{3}}} [/tex]. Created by egma2005pbjks6. matematika-ru

1) неверное при любом k∈Z2) верное при любом k∈ZОтвет: 0,5+6k; k∈Z

tex]...

egma2005pbjks6 @egma2005pbjks6

July 2022 1 4 Report

[tex] 5^\frac12 + 5^{\frac12+log_{5}sin{\frac{ \pi x}{3}}} = 15^{\frac12+log_{15}cos{\frac{ \pi x}{3}}} [/tex]

Answers & Comments

artalex74

Verified answer

$\begin {cases} sin\frac{\pi x}{3} >0 \\ cos\frac{\pi x}{3} >0 \end {cases} \Rightarrow 2\pi k$

$\sqrt{5} (1+sin\frac{\pi x}{3} )=\sqrt{15}cos \frac{\pi x}{3}\\ 1+sin\frac{\pi x}{3}=\sqrt{3}cos \frac{\pi x}{3}\\ \sqrt{3}cos \frac{\pi x}{3}-sin\frac{\pi x}{3}=1\\ \frac{\sqrt{3}}{2} cos \frac{\pi x}{3}-\frac{1}{2} sin\frac{\pi x}{3}= \frac{1}{2}\\ cos(\frac{\pi x}{3}+\frac{\pi}{6}) = \frac{1}{2}\\ \frac{\pi x}{3}+\frac{\pi}{6} = б\ \frac{\pi}{3} +2\pi k\\ x=-\frac{1}{2} б\ 1 +6k\\ x_1=-1,5+6k;\ x_2=0,5+6k;\ k \in Z$

1) $6k < -1,5 + 6k < 3 + 6k \Rightarrow 0$

неверное при любом k∈Z

2) $6k$

верное при любом k∈Z

Ответ: 0,5+6k; k∈Z

3 votes Thanks 3

Answers & Comments

Verified answer

рекомендуемые вопросы

Helpful Links

Helpful Social