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banzay111
@banzay111
July 2022
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Решить тригонометрическое уравнение
cos^2(x)+sinx-1=0
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axmva
cos^2(x)=1-sin^2x
1-sin^2x+sinx-1=0
-sin^2x+sinx=0
sinx обозначим за t
-t^2+t=0
t^2-t=0
t(t-1)=0
t=0, t=1
sinx=0 x=пи
sinx=1 x=пи/2
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Answers & Comments
1-sin^2x+sinx-1=0
-sin^2x+sinx=0
sinx обозначим за t
-t^2+t=0
t^2-t=0
t(t-1)=0
t=0, t=1
sinx=0 x=пи
sinx=1 x=пи/2