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

1 Answer
Sep 4, 2016

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

Explanation:

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#

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

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

#cosx = 0#

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

Hopefully this helps!