How do you find all values of x in the interval [0, 2pi] in the equation #cos^3x=cosx#?

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Answer 1 · 2016-09-10T01:49:27

#0, pi/2, pi. 3/2pi and 2pi.#

Here,

#cos^3x-cos x=0#. Factorizing,

#cos x(cosx -1(cos x+1)=0#. Equating each factor to 0,

#cos x=0 to #

# x=# an odd multiple of #pi/2 in [0, 2 pi]#

#= pi/2 and 3/2pi#

#cos x = 1 to #

#x =# an even multiple of pi #in [0, 2pi]#

.
#= 0 and 2pi#

#cos x = -1 to#

# x =# an odd multiple of pi #in [0, 2pi]#

.
#= pi#.

So, the list is #0, pi/2, pi, 3/2pi and 2pi#.