How do you solve #cos^2(3x)+3sin(3x)=3# ?

1 Answer
Feb 5, 2018

#x = pi/6+(2npi)/3" "# for any integer #n#

Explanation:

Given:

#cos^2(3x)+3sin(3x)=3#

Note that by Pythagoras:

#cos^2 t + sin^2 t = 1#

So:

#3 = cos^2(3x)+3sin(3x) = (1-sin^2(3x))+3sin(3x)#

Rearranging slightly:

#0 = sin^2(3x)-3sin(3x)+2 = (sin(3x)-1)(sin(3x)-2)#

Note that for real values of #t#, #-1 <= sin t <= 1#. So no real value of #x# will result in #(sin(3x)-2) = 0#

That leaves:

#sin(3x)-1 = 0#

So:

#sin(3x) = 1#

Hence:

#3x = pi/2+2npi" "# for any integer #n#

Dividing both sides by #3#, we get:

#x = pi/6+(2npi)/3" "# for any integer #n#