Question #bfc9a

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Question #bfc9a

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Answer 1 · 2018-02-12T09:33:36

$x = 0, 2 π$ $x=0,2pi$

Explanation:

Your question is

$cos (x - \frac{π}{6}) + cos (x + \frac{π}{6}) = \sqrt{3}$ $cos(x-pi/6) + cos(x+pi/6) = sqrt3$ in the interval $[0, 2 π]$ $[0,2pi]$ .

We know from trig identities that

$cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B) = cosAcosB-sinAsinB$
$cos (A - B) = cos A cos B + sin A sin B$ $cos(A-B) = cosAcosB+sinAsinB$

so that gives

$cos (x - \frac{π}{6}) = cos x cos (\frac{π}{6}) + sin x sin (\frac{π}{6})$ $cos(x-pi/6) = cosxcos(pi/6)+sinxsin(pi/6)$
$cos (x + \frac{π}{6}) = cos x cos (\frac{π}{6}) - sin x sin (\frac{π}{6})$ $cos(x+pi/6) = cosxcos(pi/6)-sinxsin(pi/6)$

therefore,

$cos (x - \frac{π}{6}) + cos (x + \frac{π}{6})$ $cos(x-pi/6)+cos(x+pi/6)$
$= cos x cos (\frac{π}{6}) + sin x sin (\frac{π}{6}) + cos x cos (\frac{π}{6}) - sin x sin (\frac{π}{6})$ $=cosxcos(pi/6)+sinxsin(pi/6)+cosxcos(pi/6)-sinxsin(pi/6)$
$= 2 cos x cos (\frac{π}{6})$ $=2cosxcos(pi/6)$

So we now know we can simplify the equation to

$2 cos x cos (\frac{π}{6}) = \sqrt{3}$ $2cosxcos(pi/6) = sqrt3$

$cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $cos(pi/6) = sqrt3/2$

so

$\sqrt{3} cos x = \sqrt{3} \to cos x = 1$ $sqrt3cosx = sqrt3 -> cosx = 1$

We know that in the interval $[0, 2 π]$ $[0,2pi]$ , $cos x = 1$ $cosx=1$ when $x = 0, 2 π$ $x=0, 2pi$

Answer 2 · 2018-02-12T09:39:24

$No soln. in (0, 2 π)$ $"No soln. in "(0,2pi)$ .

Explanation:

$cos (x - \frac{π}{6}) + cos (x + \frac{π}{6}) = \sqrt{3}$ $cos(x-pi/6)+cos(x+pi/6)=sqrt3$

Using, $cos C + cos D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2})$ $cosC+cosD=2cos((C+D)/2)cos((C-D)/2)$ ,

$2 cos x cos (- \frac{π}{6}) = \sqrt{3}$ $2cosxcos(-pi/6)=sqrt3$ ,

$:. 2*sqrt3/2*cosx=sqrt3$ ,

$:. cosx=1=cos0$ .

Now, $cosx=cosy rArr x=2kpi+-y, k in ZZ$ .

$:. cosx=cos0 rArr x=2kpi, k in ZZ, i.e.,$

$x=0,+-2pi, +-4pi,...$

$:." The Soln. Set" sub (0,2pi)" is "phi$ .

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