How do you solve #2cos(3x + pi/4) =1#?

1 Answer
A. S. Adikesavan
Jun 3, 2016

The general solution is #(24n+1)pi/36 and (24n-7)pi/36, n = 0, 1, 2, 3,...#,
#n=0: pi/36 in (0, pi/2) and -7/36pi in (-pi/2, 0). #The solution #in (0, pi)# are #pi/36 and (25/36)pi#.

Explanation:

#cos(3x+pi/4)=1/2=cos(+-pi/3,)# and, for the general solution,

#cos(3x+pi/4)=1/2=cos(2npi+-pi/3), n=0, 1, 2, ..#

So, the general solution is

#3x+pi/4=2npi+-pi/3, n=0, 1, 2, ..#. Solving,

#x=(24n+1)pi/36 and x=(24n-7)pi/36, n = 0, 1, 2, 3,...#,

The solution #in (0, pi)# are #pi/36 and (25/36)pi#.

The second comes from n=1 of #x=(24n+1)pi/36#.

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