Я повееешууусь.......... решите плииииз Sin^4x/4-cos^4x/4=1/2. Created by elizabeth1atk. matematika-ru

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July 2022 1 6 Report

Я повееешууусь.......... решите плииииз
Sin^4x/4-cos^4x/4=1/2

Answers & Comments

zuzylapa13 5* 2tg (a/2) / (1+ tg^2 (a/2)) + (1-tg^2 (a/2)/ (1+tg^2 (a/2) ) = 5
10tg (a/2) + 1- tg^2 (a/2) = 5 +5tg^2 (a/2)
6tg^2 (a/2)-10tg(a/2)-4=0
tg(a/2) = x
6x^2 - 10x-4=0
3x^2 - 5x - 2=0
d=25+4*3*2=49
x1=2
x2=-1/3
tg (a/2)=2
a/2= arctg2+Pi*R
a1=2arctg2+2Pi*R
tg(a/2)=-1/3
a2=2arctg(-1/3)+2Pi*R

2) sin^4 x + cos^4 x = sin2x - 1/2
(cos 4x - 4cos2x +3)/8 + (cos 4x + 4cos2x +3)/8=sin2x - 1/2
2cos4x +6=(sin2x - 1/2)*8
2(1-2sin^2 2x) + 6 = 8sin2x - 4
sin^2 2x + 2sin2x - 3=0
a^2 + 2a -3 =0
D=4+4*3=16
a1=1
a2=-3 (no)
sin2x=1
2x= Pi+2Pi*R

1 votes Thanks 1

elizabeth1atk Спасииибоооо большоооееееее!!!!

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