Задача 32341 tg(2pi-x)cos(3pi/2+2x)=sin(-pi/2) [2pi;7pi/2]...

shkolonik

2018-12-23 16:10:13

tg(2pi-x)cos(3pi/2+2x)=sin(-pi/2)
[2pi;7pi/2]

sova

2018-12-23 16:46:41

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tg(2π–x)=-tgx
cos(3π/2+2x)=sin2x
sin(–π/2)=-1

-tgx*(2sinx*cosx)=-1

cosx≠ 0

2sin^2x=1
sinx= ± 1/sqrt(2)

x= ±(π/4)+πk, k ∈ Zπ

б)(π/4)+2π=9π/4

(3π/4)+2π=11π/4

(5π/4)+2π=13π/4