Question

Решить 13 номер из егэ 49^sinx=(1/7)^-√2sin2x [2pi; 7pi/2]

Answer 1

$\displaystyle 49^{sinx}=( \frac{1}{7})^{- \sqrt{2}sin2x}$

$\displaystyle 7^{2sinx}=7^{ \sqrt{2}sin2x}$

$\displaystyle 2sinx= \sqrt{2}sin2x 2sinx- \sqrt{2}*2sinx*cosx=0 2sinx(1- \sqrt{2}cosx)=0$

$\displaystyle sinx=0 x= \pi n; n\in Z$

$\displaystyle 1- \sqrt{2}cosx=0$

$\displaystyle cosx= \frac{1}{ \sqrt{2}}$

$\displaystyle x=+/- \frac{ \pi }{4}+2 \pi n; n\in Z$

на промежутке [2π;7π/2]

x=2π; 3π; 9π/4