Решите пжл уравнение!!срочно!
[tex]cos^4(\pi-\frac{x}{4})+sin^2\frac{x}{4}=1[/tex]
сos^4(п-x/4)+sin^2(x/4)=1cos^4(x/4)+sin^2(x/4)=1(1-sin^2(x/4))^2 +sin^2(x/4)=11-2sin^2(x/4)+sin^4(x/4) +sin^2(x/4)=1sin^4(x/4)-sin^2(x/4)=0sin^2(x/4)*(sin^2(x/4)-1)=0sin^2(x/4)=0 sin^2(x/4)=1sin(x/4)=0 (1-cos(x/2)/2=1x/4=пk 1-cos(x/2)=2x=4пk cos(x/2)=-1 x/2=п+2пk x=2п+2пk Объединяя эти два решения: x=2пk
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Answers & Comments
сos^4(п-x/4)+sin^2(x/4)=1
cos^4(x/4)+sin^2(x/4)=1
(1-sin^2(x/4))^2 +sin^2(x/4)=1
1-2sin^2(x/4)+sin^4(x/4) +sin^2(x/4)=1
sin^4(x/4)-sin^2(x/4)=0
sin^2(x/4)*(sin^2(x/4)-1)=0
sin^2(x/4)=0 sin^2(x/4)=1
sin(x/4)=0 (1-cos(x/2)/2=1
x/4=пk 1-cos(x/2)=2
x=4пk cos(x/2)=-1
x/2=п+2пk
x=2п+2пk
Объединяя эти два решения: x=2пk