How do you solve $cos 3 x = 0$ $cos3x=0$ over the interval 0 to 2pi?

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Answer 1 · 2016-09-04T22:01:49

$x = \frac{π}{2} and \frac{3 π}{2}$ $x = pi/2 and (3pi)/2$

Apply the identity $cos (3 x) = cos (2 x + x)$ $cos(3x) = cos(2x + x)$ :

$cos (2 x + x) = 0$ $cos(2x + x ) = 0$

We can now expand using the sum formula $cos (A + B) = cos A cos B - sin A sin B$ $cos(A + B)= cosAcosB - sinAsinB$ .

$cos 2 x cos x + sin 2 x sin x = 0$ $cos2xcosx + sin2xsinx = 0$

Use the following double angle identities to expand further:

• $cos 2 α = 1 - 2 {sin}^{2} α$ $cos2alpha = 1 - 2sin^2alpha$

• $sin 2 α = 2 sin α cos α$ $sin2alpha = 2sinalphacosalpha$

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$(1 - 2 {sin}^{2} x) cos x + 2 sin x cos x sin x = 0$ $(1 - 2sin^2x)cosx + 2sinxcosxsinx = 0$

$cos x - 2 {sin}^{2} x cos x + 2 {sin}^{2} x cos x = 0$ $cosx - 2sin^2xcosx + 2sin^2xcosx = 0$

$cos x = 0$ $cosx = 0$

$x = \frac{π}{2} and \frac{3 π}{2}$ $x = pi/2 and (3pi)/2$

Hopefully this helps!

How do you solve cos3x=0cos3x=0cos3x=0 over the interval 0 to 2pi?