How do you solve $cos (2 x) cos (x) - sin (2 x) sin (x) = 0$ $Cos(2x)cos(x)-sin(2x)sin(x)=0$ ?

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How do you solve $cos (2 x) cos (x) - sin (2 x) sin (x) = 0$ $Cos(2x)cos(x)-sin(2x)sin(x)=0$ ?

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Answer 1 · 2018-05-27T19:21:40

We have:

$cos (2 x) cos x = sin (2 x) sin x$ $cos(2x)cosx = sin(2x)sinx$

$1 = tan (2 x) tan x$ $1 = tan(2x)tanx$

Now we have to make some manipulations using $tan (2 x) = \frac{2 sin x cos x}{{cos}^{2} x - {sin}^{2} x}$ $tan(2x) = (2sinxcosx)/(cos^2x- sin^2x)$

$1 = \frac{2 sin x cos x}{{cos}^{2} x - {sin}^{2} x} \cdot \frac{sin x}{cos x}$ $1 = (2sinxcosx)/(cos^2x- sin^2x) * sinx/cosx$

${cos}^{2} x - {sin}^{2} x = 2 {sin}^{2} x$ $cos^2x -sin^2x = 2sin^2x$

$1 - {sin}^{2} x - {sin}^{2} x = 2 {sin}^{2} x$ $1 - sin^2x - sin^2x = 2sin^2x$

$1 = 4 {sin}^{2} x$ $1 = 4sin^2x$

$\frac{1}{4} = {sin}^{2} x$ $1/4 = sin^2x$

$sin x = \pm \frac{1}{2}$ $sinx = +- 1/2$

It's now clear that $x = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6} and \frac{11 π}{6}$ $x = pi/6, (5pi)/6, (7pi)/6 and (11pi)/6$ . In general form, these are $\frac{π}{6} + π n$ $pi/6 + pin$ and $\frac{5 π}{6} + π n$ $(5pi)/6 + pin$ . However we must also include when $cos x = 0$ $cosx = 0$ , namely when $x = \frac{π}{2} + π n$ $x = pi/2 + pin$ . We can confirm graphically.

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Hopefully this helps!

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How do you solve $cos (2 x) cos (x) - sin (2 x) sin (x) = 0$ $Cos(2x)cos(x)-sin(2x)sin(x)=0$ ?

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