Verified Answer :

caylus

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:

Given,

\(8 cos^3 (x+\frac{\pi}{3})=cos 3x\)

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

Also, cos 3A = 4cos³ A - 3 cos A,

\(\implies 8 (cos x cos(\frac{\pi}{3}) - sin x sin (\frac{\pi}{3}) )^3 = 4cos^3 x - 3cos x\)

\(8(cosx\times \frac{1}{2} - sin x\times \frac{\sqrt{3}}{2})^3 = 4cos^3 x - 3cosx\)

\((cos x-\sqrt{3} sinx)^3 = 4cos^3 x - 3 cosx\)

\(cos^3 x-3\sqrt{3} sin^3 x-3\sqrt{3} sin x cos^2 x + 9 sin^2x cos x = 4cos^3 x - 3 cosx\)

\(cos^3x - 3\sqrt{3} sinx ( sin^2x + cos^2x) + 9 sin^2x cos x = 4cos^3 x - 3 cosx\)

\(cos^3x - 3\sqrt{3} sin x + 9 sin^2x cos x - 4 cos^3x + 3 cosx\)

\(-3cos^3x - 3\sqrt{3} sin x + 9 sin^2x cos x + 3cos x = 0\)

\(cos^3x +\sqrt{3} sin x - 3 sin^2x cos x - cos x = 0\)

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

( ∵ sin²x = 1 - cos² x )

\(-sin^2x cosx +\sqrt{3} sin x - 3 sin^2x cos x=0\)

\(-4sin^2x cosx +\sqrt{3} sin x =0\)

\(-2sin x sin 2x + \sqrt{3} sin x =0\)

\(sin x (-2sin 2x + \sqrt{3})=0\)

By zero product property,

sin x = 0 or -2 sin 2x + √3 = 0

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

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

Where, n is an integer,

Related Questions