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KnyazSlavaGalicskii
@KnyazSlavaGalicskii
August 2022
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Решить тригонометриченское уравнение .
2 tg²x + 3 tgx - 2 = 0
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evgennik3
Пусть tg x=t
2*t^2+3t-2=0
D=b^2-4ac=3^2-4*2*(-2)=5^2
t(1,2)=(-b
±√D)/2a
t1=(-3+5)/(2*2)=0.5 t2=(-3-5)/4=-2
tg x = t1 и tg x =t2
tg x = 0.5 tg x = -2
x= arctg(0.5)+пи*k, k
∈Z x=arctg(-2)+пи*k, k∈Z
x= -arctg 2 +пи*k, k∈Z
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Answers & Comments
2*t^2+3t-2=0
D=b^2-4ac=3^2-4*2*(-2)=5^2
t(1,2)=(-b±√D)/2a
t1=(-3+5)/(2*2)=0.5 t2=(-3-5)/4=-2
tg x = t1 и tg x =t2
tg x = 0.5 tg x = -2
x= arctg(0.5)+пи*k, k∈Z x=arctg(-2)+пи*k, k∈Z
x= -arctg 2 +пи*k, k∈Z