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вор9
вор9
01.05.2020 09:31 •  Алгебра

Объясните как решать эти примеры: 1)2sin^2(x)+cos4x=0 2) 5sin^2(x)-cos^2(x)=4+4sinxcosx

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Ответ:
RedBalloon
RedBalloon
01.05.2020
1. Используем 2 преобразования;
2*(sin(x))^2=1-cos(2x) и 
cos(4x)= 2*cos(2x)^2 - 1 ⇒ 2*cos(2x)^2=cos(2x) ⇒⇒
cos(2x)=0 ⇒ x=(+/-)π/4+π*n 
cos(2x)=1/2 ⇒ x=(+/-)π/12+π*n
4,6(47 оценок)
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Ответ:
gulmamedovaxayald
gulmamedovaxayald
01.05.2020
X = π/3 - y
sin(π/3 - y)*siny = 1/4
(sin(π/3)*cosy - siny*cos(π/3) )*siny = 1/4
((cosy)*√3/2 - (siny)/2)*siny = 1/4
(cosy*siny*√3 - sin^2(y))/2 = 1/4
√3*cosy*siny - sin^2(y) = 1/2
1/2 = 0.5sin^2(y) + 0.5cos^2(y)
√3*cosy*siny - sin^2(y) - 0.5sin^2(y) - 0.5cos^2(y) = 0 - делим на -0.5
cos^2(y) + 3sin^2(y) - 2√3*cosy*siny = 0 - делим на cos^2(y)
1 + 3tg^2(y) - 2√3*tgy = 0
замена tg(y) = t
3t^2 - 2√3*t + 1 = 0
(√3t - 1)^2 = 0
√3t = 1, t = √3/3
tg(y) = √3/3
y = π/3 + πk
x = π/3 - π/3 - πk = -πk
ответ: x = -πk, y = π/3 + πk
4,8(99 оценок)
Ответ:
Angel168Demon
Angel168Demon
01.05.2020

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4,5(20 оценок)
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