Question

\(\cos^3x-4\sin^3x-3\cos x\sin^2x+\sin x=0\)

Answer 1

+ , cos³ x = 0 => 0 - 4 - 0 + 1 = 0 ( vô nghiệm)

+, cos³ x \(\ne\)0 , chia cả 2 vế của pt cho cos³ x , ta đc

\(\frac{\cos^3x-4sin^3x-3cosx.sin^2x+sinx}{cos^3x}=0\)

1 - \(\frac{4\sin^3x}{\cos^3x}\) - \(\frac{3\sin^2x}{cos^2x}\) + \(\frac{1}{\cos^2x}\)= 0

1 - 4 tan³x - 3 tan²x + 1 + tan²x = 0

-4 tan³x - 2tan²x + 2 = 0

=> tan x = tan \(\alpha\) ( tan \(\alpha\approx0,66\))

=> x = \(\alpha+k.\pi\)