The number of solution of the equation cosx cos2x cos3x=1/4?

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The number of solution of the equation cosx cos2x cos3x=1/4?

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Answer 1 · 2018-01-14T11:18:38

There are six solutions between any two cobsecutive integer multiples of $π$ $\pi$ . See below.

Explanation:

Note first that

$cos (x + π) = - cos (x)$ $\cos(x+\pi)=-\cos(x)$

$cos (2 (x + π)) = cos (2 x)$ $\cos(2(x+\pi))=\cos(2x)$

$cos (3 (x + π)) = - cos (3 x)$ $\cos(3(x+\pi))=-\cos(3x)$

So the function $f (x) = cos (x) cos (2 x) cos (3 x)$ $f(x)=\cos(x)\cos(2x)\cos(3x)$ has a period $π$ $\pi$ , and its solutions for a value of $\frac{1}{4}$ $1/4$ will follow suit.

Using the sum product relation for cosines:

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

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

So then

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

$2 (cos (2 x) + cos (4 x)) = 1$ $2(\cos(2x)+\cos(4x))=1$

Then from the double angle formula:

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

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

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

Note that the equation comes out with trigonometric functions of $2 x$ $2x$ because of the period being $π$ $\pi$ .

We can factor the zero polynomial:

$(2 cos (2 x) + 1) (2 {cos}^{2} (2 x) - 1) = 0$ $(2\cos(2x)+1)(2\cos^2(2x)-1)=0$

So either $cos (2 x) = - (\frac{1}{2})$ $\cos(2x)=-(1/2)$ or else $cos (2 x) = \pm (\frac{\sqrt{2}}{2})$ $\cos(2x)=\pm(\sqrt(2)/2)$ . This leads to six sets of solutions having period $π$ $\pi$ :

If $cos (2 x) = - (\frac{1}{2})$ $\cos(2x)=-(1/2)$ then $x = \pm (\frac{π}{3}) + n π$ $x=\pm(\pi/3)+n\pi$ , for any integer $n$ $n$ .

If $cos (2 x) = (\frac{\sqrt{2}}{2})$ $\cos(2x)=((\sqrt(2))/2)$ then $x = \pm (\frac{π}{8}) + n π$ $x=\pm(\pi/8)+n\pi$ , for any integer $n$ $n$ .

If $cos (2 x) = (- \frac{\sqrt{2}}{2})$ $\cos(2x)=(-(\sqrt(2))/2)$ then $x = \pm (\frac{3 π}{8}) + n π$ $x=\pm((3\pi)/8)+n\pi$ , for any integer $n$ $n$ .

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The number of solution of the equation cosx cos2x cos3x=1/4?

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