Question #12b1c

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Question #12b1c

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Answer 1 · 2017-12-21T21:10:50

The solutions are $\frac{k}{18} π$ $k/18pi$ , where $k \in {0, 5, 7, 17, 18, 19, 29, 31}$ $k in {0,5,7,17,18,19,29,31}$

Explanation:

$2 tan (x) cos (3 x) = - \sqrt{3} tan (x)$ $2tan(x)cos(3x)=-sqrt(3)tan(x)$

Move everything to one side
$2 tan (x) cos (3 x) + \sqrt{3} tan (x) = 0$ $2tan(x)cos(3x)+sqrt(3)tan(x)=0$

Factor $tan (x)$ $tan(x)$ out
$tan (x) (2 cos (3 x) + \sqrt{3}) = 0$ $tan(x)(2cos(3x)+sqrt(3))=0$

By zero product property, we have
$tan (x) = 0$ $tan(x)=0$ or $2 cos (3 x) + \sqrt{3} = 0$ $2cos(3x)+sqrt(3)=0$

$tan (x) = 0$ $tan(x)=0$ has solutions $x = k π$ $x=kpi$ , where $k$ $k$ is an integer.
Since $x \in [0, 2 π)$ $x in [0,2pi)$ , only 0 and $π$ $pi$ are solutions.

Moving on with the other term
$2 cos (3 x) + \sqrt{3} = 0$ $2cos(3x)+sqrt(3)=0$
$cos (3 x) + \frac{\sqrt{3}}{2} = 0$ $cos(3x)+sqrt(3)/2=0$
$cos (3 x) = - \frac{\sqrt{3}}{2}$ $cos(3x)=-sqrt(3)/2$
Here a graph of $cos (x)$ $cos(x)$ would be helpful - it gives a little suggestion (click and drag for (x,y) values and special points)
graph{(y-cos(x))(y+sqrt(3)/2)=0 [-1,7,-2,2]}
We need to recall, that $\frac{\sqrt{3}}{2} = cos (\frac{π}{6})$ $sqrt(3)/2=cos(pi/6)$
From the graph we see, that the solution is $3 x = 2 k π + π \pm \frac{π}{6}$ $3x=2kpi+pi pm pi/6$ or $x = (\frac{2 k + 1}{3} \pm \frac{1}{18}) π$ $x=((2k+1)/3pm1/18)pi$
Since $x \in [0, 2 π)$ $x in [0,2pi)$ , only solutions are
$(\frac{1}{3} - \frac{1}{18}) π, (\frac{1}{3} + \frac{1}{18}) π, (\frac{3}{3} - \frac{1}{18}) π, (\frac{3}{3} + \frac{1}{18}) π, (\frac{5}{3} - \frac{1}{18}) π, (\frac{5}{3} + \frac{1}{18}) π$ $(1/3-1/18)pi,(1/3+1/18)pi,(3/3-1/18)pi,(3/3+1/18)pi,(5/3-1/18)pi,(5/3+1/18)pi$
or simplified
$\frac{5}{18} π, \frac{7}{18} π, \frac{17}{18} π, \frac{19}{18} π, \frac{29}{18} π, \frac{31}{18} π$ $5/18pi,7/18pi,17/18pi,19/18pi,29/18pi,31/18pi$

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Question #12b1c

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