How do you find all solutions of the equation in the interval #[0,2pi)# given #cos4x(cosx-1)=0#?

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How do you find all solutions of the equation in the interval #[0,2pi)# given #cos4x(cosx-1)=0#?

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Answer 1 · 2018-07-09T14:22:20

#x=pi/8,3*pi/8,5pi/8,7pi/8,9pi/8,11pi/8,13pi/8,15pi/8# or #x=0#

Explanation:

We have to distinguish two cases:
#cos(4x)=0#
this gives
#4x=(2k+1)*pi/2#
#x=(2k+1)*pi/8#
the interval for #k# is #1\le k\le 7# and #k# is an integer.

#cos(x)=1#

here is only #x=0# one Solutions.

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How do you find all solutions of the equation in the interval #[0,2pi)# given #cos4x(cosx-1)=0#?

1 Answer

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