Question

cosx+cos2x+cos3x+cos4x=0 number of real values of x?

Answer 1

I hope this will help u...

Answer 2

Answer:

$x=(2n+1)(\frac{\pi}{2})$

$x=(2n+1)(\frac{\pi}{5})$

$x=(2n+1)\pi$

Step-by-step explanation:

Given : Expression $\cos x+\cos2x+\cos3x+\cos4x=0$

To find : Number of real values of x?

Solution :

Re-write expression as $\cos x+\cos3x+\cos2x+\cos4x=0$

Applying trigonometric identity,

$\cos A+\cos B =2\cos (\frac{A+B}{2})\cos (\frac{A-B}{2})$

$2\cos (\frac{x+3x}{2})\cos (\frac{x-3x}{2})+2\cos(\frac{2x+4x}{2})\cos (\frac{2x-4x}{2})=0$

$2\cos 2x\cos x+2\cos3x\cos x=0$

$2\cos x(\cos 2x+\cos3x)=0$

Applying again,

$2\cos x(2\cos (\frac{2x+3x}{2})\cos (\frac{2x-3x}{2}))=0$

$2\cos x(2\cos (\frac{5x}{2})\cos (\frac{x}{2}))=0$

$\cos x\times \cos (\frac{5x}{2})\times \cos (\frac{x}{2})=0$

i.e. $\cos x=0$

$\cos x=\cos (\frac{\pi}{2})$

$x=(2n+1)(\frac{\pi}{2})$

or $\cos (\frac{5x}{2})=0$

$\cos (\frac{5x}{2})=\cos (\frac{\pi}{2})$

$\frac{5x}{2}=(2n+1)(\frac{\pi}{2})$

$x=(2n+1)(\frac{\pi}{5})$

or $\cos (\frac{x}{2})=0$

$\cos (\frac{x}{2})=\cos (\frac{\pi}{2})$

$\frac{x}{2}=(2n+1)(\frac{\pi}{2})$

$x=(2n+1)\pi$