2sin (pi/4-x) = sin2x-2cos^2x

Question

Алгебра

Венура

25 сентября, 13:25

2sin (pi/4-x) = sin2x-2cos^2x

+4

Ответы (1)

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Фрида · Answer 1

2sin (π/4) cosx - 2cos (π/4) sinx=2sinxcosx-2cos^2x,

2 (√2/2) cosx - 2 (√2/2) sinx=2cosx (sinx-cosx),

(√2/2) (cosx - sinx) = cosx (sinx-cosx),

cosx (sinx - cosx) + (√2/2) (sinx - cosx) = 0,

(sinx - cosx) (cosx+√2/2) = 0

1) sinx=cosx, x=π/4 + πk, k∈Z

2) cosx = - √2/2, x = + - 3π/4 + 2πk, k∈Z