Задача 37144 ...

vk254337550

2019-05-16 20:52:45

3cos2x + 1 = sin(π/2 - x)

sova

2019-05-16 20:59:05

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cos2x=2cos^2x-1

По формуле приведения:

sin((π/2) – x)=cosx

6cos^2x-3+1-cosx=0

6cos^2x-cosx-2=0

D=1+4*6*2=49

cosx=-1/2 или cosx=2/3

x= [b] ± (2π/3)+2πn, n ∈Z [/b] или х= [b]± arccos(2/3)+2πm, m ∈ Z[/b]