Решить уравнения: а)Sin4x + cos^2 2x = 2 б)cos^4(3/2)x-sin^4(3/2)x=1/2. Created by vanek20003. algebra-ru

А)Действительных корней нет.б)n∈Z

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vanek20003 @vanek20003

July 2022 1 1 Report

Решить уравнения:
а)Sin4x + cos^2 2x = 2
б)cos^4(3/2)x-sin^4(3/2)x=1/2

Answers & Comments

KayKosades А) $sin4x+cos^22x=2 \\ 
2sin2xcos2x+cos^22x-2cos^22x-2sin^22x=0 \\ 
2sin^22x-2sin2xcos2x+cos^22x=0 |:cos^22x \neq 0 \\ 
2tg^22x-2tg2x+1=0 \\ 
tg2x=t \\ 
2t^2-2t+1=0 \\ 
D\ \textless \ 0
$
Действительных корней нет.
б)
$cos^4( \frac{3x}{2} )-sin^4( \frac{3x}{2} )= \frac{1}{2} \\ 
(cos^2( \frac{3x}{2} )-sin^2( \frac{3x}{2} ))(cos^2( \frac{3x}{2} )+sin^2( \frac{3x}{2} ))= \frac{1}{2} \\ 
cos^2( \frac{3x}{2} )-sin^2( \frac{3x}{2} )= \frac{1}{2} \\ 
cos^2( \frac{3x}{2} )-(1-cos^2( \frac{3x}{2} ))= \frac{1}{2} \\ 
2cos^2( \frac{3x}{2} )= \frac{3}{2} \\ 
cos^2( \frac{3x}{2} )= \frac{3}{4} \\ 
cos( \frac{3x}{2} )= \frac{ \sqrt{3} }{2} \\ 
 \frac{3x}{2} =+- \frac{ \pi }{6} +2 \pi n \\ 
$
$x=+- \frac{ \pi }{9} + \frac{2 \pi n}{3} \\ 
cos( \frac{3x}{2} )=- \frac{ \sqrt{3}}{2} \\ 
x=+- \frac{5 \pi }{9} + \frac{2 \pi n}{3} 

$
n∈Z

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