Question

Hello i'm vietnamese help me
8cos^3 (x+pi/3)= cos 3x

Answer 1

Hello,

Remember:

1) cos 3a=4cos^3 a -3cos a
2) cos (a+b)=cos a cos b- sin a sin b
3) cos π/3=1/2
4) sin π/3=√3 /2
5) cos (x+π/3)=cos x cos π/3 - sin x sin π/3=1/2(cos x -√3 sin x)


8 cos ^3 (x+π/3)=cos 3x
==> (cos x -√3 sin x)^3 =4cos^3 x-3cos x
==> cos ^3 x -3√3cos²x sin x+9 cos x sin² x-3√3 sin^3 x -4cos^3 x+3 cos x=0
==> -3 cos^3 x -3√3 sin x(cos²x+sin²x)+9 cos x sin² x +3 cos x=0
==>3 cos ^3 x +3√3 sin x-9 cos x sin² x -3 cos x=0
==>cos ^3 x - cos x +√3 sin x -3 cos x (1-cos² x)=0
==>cos^3 x - cos x +√3 sin x -3 cos x +3 cos^3 x=0
==>4cos^3 x-4 cos x +√3 sin x=0
==>4 cos x(cos²x-1)+√3 sin x=0
==>-4 cos x sin²x+√3 sin x=0
==>sin x(-4sin x cos x +√3)=0
==>sin x(-2sin 2x+√3)=0
==> sin x=0 or sin 2x=√3/2
==> x=kπ or 2x=π/3+2kπ or 2x=2π/3+2kπ
==> x=kπ or x=π/6+kπ or x=π/3+kπ

Answer 2

Given,

`8 cos^3 (x+(\pi)/(3))=cos 3x`

∵ cos (A + B) = cos A.cos B - sin A.sin B

Also, cos 3A = 4cos³ A - 3 cos A,

`\implies 8 (cos x cos((\pi)/(3)) - sin x sin ((\pi)/(3)) )^3 = 4cos^3 x - 3cos x`

`8(cosx* (1)/(2) - sin x* (√(3))/(2))^3 = 4cos^3 x - 3cosx`

`(cos x-√(3) sinx)^3 = 4cos^3 x - 3 cosx`

`cos^3 x-3√(3) sin^3 x-3√(3) sin x cos^2 x + 9 sin^2x cos x = 4cos^3 x - 3 cosx`

`cos^3x - 3√(3) sinx ( sin^2x + cos^2x) + 9 sin^2x cos x = 4cos^3 x - 3 cosx`

`cos^3x - 3√(3) sin x + 9 sin^2x cos x - 4 cos^3x + 3 cosx`

`-3cos^3x - 3√(3) sin x + 9 sin^2x cos x + 3cos x = 0`

`cos^3x +√(3) sin x - 3 sin^2x cos x - cos x = 0`

`cosx (cos^2 - 1) + √(3) sin x - 3 sin^2x cos x=0`

( ∵ sin²x = 1 - cos² x )

`-sin^2x cosx +√(3) sin x - 3 sin^2x cos x=0`

`-4sin^2x cosx +√(3) sin x =0`

`-2sin x sin 2x + √(3) sin x =0`

`sin x (-2sin 2x + √(3))=0`

By zero product property,

sin x = 0 or -2 sin 2x + √3 = 0

`x=n\pi\text{ or }2x = (\pi)/(3)+n\pi`

`\implies x = n\pi\text{ or }x = (\pi)/(6) + (n\pi)/(2)`

Where, n is an integer,