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Санька98
@Санька98
July 2022
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(2sin 2x- cos 2x) (1+cos 2x) = sin^2 2x. решите, спасибо.
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Удачник66
Verified answer
2sin 2x - cos 2x + 2sin 2x*cos 2x - cos^2 2x = sin^2 2x
2sin 2x - cos 2x + 2sin 2x*cos 2x = sin^2 2x +cos^2 2x = 1
2sin 2x*(1 + cos 2x) = 1 + cos 2x
(1 + cos 2x)(2sin 2x - 1) = 0
1) 1 + cos 2x = 0
cos 2x = -1
2x = pi + 2pi*k
x1 = pi/2 + pi*k
2) 2sin 2x = 1
sin 2x = 1/2
2x = pi/6 + 2pi*n
x2 = pi/12 + pi*n
2x = 5pi/6 + 2pi*n
x3 = 5pi/12 + pi*n
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Verified answer
2sin 2x - cos 2x + 2sin 2x*cos 2x - cos^2 2x = sin^2 2x2sin 2x - cos 2x + 2sin 2x*cos 2x = sin^2 2x +cos^2 2x = 1
2sin 2x*(1 + cos 2x) = 1 + cos 2x
(1 + cos 2x)(2sin 2x - 1) = 0
1) 1 + cos 2x = 0
cos 2x = -1
2x = pi + 2pi*k
x1 = pi/2 + pi*k
2) 2sin 2x = 1
sin 2x = 1/2
2x = pi/6 + 2pi*n
x2 = pi/12 + pi*n
2x = 5pi/6 + 2pi*n
x3 = 5pi/12 + pi*n