How do you solve #cos^2(3x)+3sin(3x)=3# ?
1 Answer
Feb 5, 2018
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
That leaves:
#sin(3x)-1 = 0#
So:
#sin(3x) = 1#
Hence:
#3x = pi/2+2npi" "# for any integer#n#
Dividing both sides by
#x = pi/6+(2npi)/3" "# for any integer#n#