Question

Cosx+cos2x+cos3x+cos4x=-1/2 number of real values of x

Answer 1

Consider $\cos x + \cos 2x + \cos 3x + \cos 4x = \frac{-1}{2}$

$\cos x + \cos 3x + \cos 2x + \cos 4x = \frac{-1}{2}$

Now, using the trigonometric identity $\cos A + \cos B = 2 \cos (\frac{A-B}{2}) \cos (\frac{A+B}{2})$

$2 \cos (\frac{x-3x}{2}) \cos (\frac{x+3x}{2})+2 \cos (\frac{2x-4x}{2}) \cos (\frac{2x+4x}{2}) = \frac{-1}{2}$

$2 \cos (-x) \cos (2x)+2 \cos (-x) \cos (3x)= \frac{-1}{2}$

$2 \cos x \cos 2x+2 \cos x \cos 3x= \frac{-1}{2}$

$2 \cos x ( \cos 2x+ \cos 3x )= \frac{-1}{2}$

$2 \cos x ( 2\cos \frac{x}{2} \cos \frac{5x}{2} )= \frac{-1}{2}$

$4 \cos x ( \cos \frac{x}{2} \cos \frac{5x}{2} )= \frac{-1}{2}$

$\cos x ( \cos \frac{x}{2} \cos \frac{5x}{2} )= \frac{-1}{8}$

Multiplying both sides of the equation by $\sin{\frac{x}{2}}$

$\sin \frac{x}{2} \cos x \cos \frac{x}{2} \cos \frac{5x}{2}= \frac{-1}{8} \sin \frac{x}{2}$

$2\sin \frac{x}{2} \cos x \cos \frac{x}{2} \cos \frac{5x}{2} = \frac{-1}{4} \sin \frac{x}{2}$

$2\sin \frac{x}{2} \cos \frac{x}{2} \cos x \cos \frac{5x}{2} = \frac{-1}{4} \sin \frac{x}{2}$

$\sin x \cos x \cos \frac{5x}{2} = \frac{-1}{4} \sin \frac{x}{2}$

$2\sin x \cos x \cos \frac{5x}{2} = \frac{-1}{2} \sin \frac{x}{2}$

$\sin 2x \cos \frac{5x}{2} = \frac{-1}{2} \sin \frac{x}{2}$

$2\sin 2x \cos \frac{5x}{2} =- \sin \frac{x}{2}$

$\sin \frac{9x}{2} - \sin \frac{x}{2} =- \sin \frac{x}{2}$

$\sin \frac{9x}{2} = 0$

So, $\frac{9x}{2} = n \pi$

$9x = 2n \pi$

$x = \frac{2n \pi}{9}$ where 'n' is a natural number.