How do you solve #1=8cos(2x +1) -3#?

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Answer 1 · 2015-05-13T20:04:09

First add 3 to both sides to get:

#4 = 8 cos(2x+1)#

Next, divide both sides by 8 to get:

#1/2 = cos(2x+1)#

If #cos(2x+1) = 1/2# then #2x+1 = 2n pi +- pi/3# where #n# is any integer.

Subtract 1 from both sides to get:

#2x = 2n pi +- pi/3 - 1#

Divide both sides by 2 to get:

#x = n pi +- pi/6 - 1/2#