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

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Answer 1 · 2015-04-27T16:34:05

Solve #y = 2. cos 3(x - Pi/4) = 0 #.

Call #(x - Pi/4) = t -> y = 2. cos 3t = 0#

#cos 3t = 0 --> 3t = Pi/2; and 3t = 3Pi/2#

#3t = Pi/2 -> t = Pi/6 (1)#

#3t = 3Pi/2 -> t = Pi/2 (2)#

#(1) t = (x - Pi/4) = Pi/6 -> x = 5Pi/12#

#(2) t = (x - Pi/4) = Pi/2 -> x = Pi/2 + pi/4 = 6Pi/8 = 3Pi/4#

Answer:

#x = 5Pi/12 and x = 3Pi/4#

Check:

(1) #x = 5Pi/12 -> t = x - Pi/4 = 5Pi/12 - Pi/4 = Pi/6 -> 3t = 3Pi/6 = Pi/2 -> cos 3t = cos Pi/2 = 0#.

Correct.

(2) #x = 3Pi/4 -> t = (x - Pi/4) = 3Pi/4 - Pi/4 = Pi/2 -> 3t = 3Pi/2 -> cos 3t = cos 3Pi/2 = 0#.

Correct.