Question

2. Determine todos os n´umeros reais x tais que
2cos3x − cosx = 0

Answer 1

Resposta:

Explicação passo a passo:

cos3x =

(cos2x + x) =

cos2x.cosx - sen2x.senx =

cos2x.cosx - 2senx.cosx.senx =

(cos²x - sen²x)cosx - 2sen²x.cosx =

cos³x - sen²x.cosx - 2sen²x.cosx =

cos³x - 3sen²x.cosx

==//==

2(cos³x - 3sen²x.cosx) − cosx = 0

2cos³x - 6sen²x.cosx − cosx = 0

cosx(2cos²x - 6sen²x - 1) = 0

cosx[2cos² - 6(1-cos²x) - 1] = 0

cosx[2cos² - 6+6cos²x - 1] = 0

cosx[8cos² - 7] = 0

cosx =0. Logo x = π/2 + kπ

8cos²x - 7 = 0

cos²x = 7/8

cosx = ±√7/√8

x = arc cos √7/√8