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

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Answer 1 · 2016-09-04T22:01:49

#x = pi/2 and (3pi)/2#

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

#cos(2x + x ) = 0#

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

#cos2xcosx + sin2xsinx = 0#

Use the following double angle identities to expand further:

•#cos2alpha = 1 - 2sin^2alpha#

•#sin2alpha = 2sinalphacosalpha#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#(1 - 2sin^2x)cosx + 2sinxcosxsinx = 0#

#cosx - 2sin^2xcosx + 2sin^2xcosx = 0#

#cosx = 0#

#x = pi/2 and (3pi)/2#

Hopefully this helps!