Solve the equation #cos3t=cos2t#?

#cos2theta=cos3theta.#

# rArr cos3theta-cos2theta=0.#

# rArr -2sin((3theta+2theta)/2)*sin((3theta-2theta)/2)=0.#

# rArr -2sin((5theta)/2)sin(theta/2)=0.#

# rArr sin((5theta)/2)=0, or, sin(theta/2)=0.#

Knowing that, #sinx=o rArr x=kpi, k in ZZ,# we have,

#5/2theta=kpi, or, theta/2=kpi, k in ZZ.#

#theta=2/5kpi, or, theta=2kpi, k in ZZ.#

# rArr theta in {2/5kpi}uu{2kpi}, k in ZZ.#

But, we observe that, #{2kpi : k in ZZ} sub {2/5kpi : k in ZZ},# hence

#"The Soln. Set is "{2/5kpi : k in ZZ}.#

Enjoy Maths.!

We will use here #cosB-cosA=2sin((A+B)/2)sin((A-B)/2)#

Hence, we ccan write #cos2theta=cos3theta# as

#cos2theta-cos3theta=0#

or #2sin((3theta+2theta)/2)sin((3theta-2theta)/2)=0#

or #2sin((5theta)/2)sin(theta/2)=0#

i.e. either #(5theta)/2=npi# or #theta/2=npi#, where #n# is an integer.

i.e. either #theta=(2npi)/5# or #theta=2npi#, where #n# is an integer.

Observe that the latter set (i.e. where #theta=2npi#) is a subset of former set (i.e. where #theta=(2npi)/5#). This is so when #n# is a multiple of #5#.

Hence solution is #theta=(2npi)/5#, where #n# is an integer.

cos 3t = cos 2t
#3t = +- 2t#

a. 3t = 2t
t = 0 or #t = 2kpi#

b. 3t = - 2t
5t = 0 or #5t = 2kpi#
#t = (2kpi)/5#
Check by calculator, with k = 1
#t = (2pi)/5 = 72^@# --> #cos 2t = cos 144^@ = - 0.809# -->
#cos 3t = cos 216^@ = - 0.809#. OK
#t = 2pi# --> #cos 2t = cos 4pi = 1# and #cos 3t = cos 6pi = 1#. OK