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Eeeerokkkk
@Eeeerokkkk
July 2022
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ShirokovP
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Sin (pi/2 + 2x) + cos (pi/2 - 2x) = 0
cos 2x + sin2x = 0
cos^2x - sin^2x + 2sinxcosx = 0
- sin^2x + 2sinxcosx + cos^2x = 0
sin^2x - 2sinxcosx - cos^2x = 0 /:cos^2x ≠0
tg^2x - 2tgx - 1 = 0
tgx = t
t^2 - 2t - 1 = 0
D = 4 + 4 = 8
t1 = (2 + 2√2)/2 = 1 + √2
t2 = (2 - 2√2)/2 = 1 - √2
tgx = 1 + √2
x = arctg(1 + √2) + pik, k ∈Z
tgx = 1 - √2
x = arctg(1 - √2) + pik, k ∈Z
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Answers & Comments
Verified answer
Sin (pi/2 + 2x) + cos (pi/2 - 2x) = 0cos 2x + sin2x = 0
cos^2x - sin^2x + 2sinxcosx = 0
- sin^2x + 2sinxcosx + cos^2x = 0
sin^2x - 2sinxcosx - cos^2x = 0 /:cos^2x ≠0
tg^2x - 2tgx - 1 = 0
tgx = t
t^2 - 2t - 1 = 0
D = 4 + 4 = 8
t1 = (2 + 2√2)/2 = 1 + √2
t2 = (2 - 2√2)/2 = 1 - √2
tgx = 1 + √2
x = arctg(1 + √2) + pik, k ∈Z
tgx = 1 - √2
x = arctg(1 - √2) + pik, k ∈Z