How do you find all #sin^2 3x-sin^2x=0# in the interval #[0,2pi)#?

2 Answers
Narad T.
Jul 15, 2017

The solutions are #S={0,1/4pi,1/2pi,3/4pi,pi,5/4pi,3/2pi,7/4pi}#

Explanation:

We need

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

#sin2x=2sinxcox#

#cos2x=1-2sin^2x#

We start by calculating #sin3x#

#sin3x=sin(2x+x)#

#=sin2xcosx+cos2xsinx#

#=2sinxcos^2x+sinx(1-2sin^2x)#

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

#=2sinx-2sin^3x+sinx-2sin^3x#

#=3sinx-4sin^3x#

Therefore,

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

#(3sinx-4sin^3x)^2-sin^2x=0#

#9sin^2x-24sin^4x+16sin^6x-sin^2x=0#

#16sin^6x-24sin^4x+8sin^2x=0#

#8sin^2x(2sin^4x-3sin^2x+1)=0#

#8sin^2x(2sin^2x-1)(sin^2x-1)=0#

#8sin^2x(sqrt2sinx+1)(sqrt2sinx-1)(sinx+1)(sinx-1)=0#

So,

#sin^2x=0#, #=>#, #x=2pin#, and #x=pi+2pin#

#sqrt2sinx+1=0#, #=>#, #sinx=-1/sqrt2#, #=>#, #x=5/4pi+2pin#, and #x=7/4pi+2pin#

#sqrt2sinx-1=0#, #=>#, #sinx=1/sqrt2#, #=>#, #x=1/4pi+2pin# and #x=3/4pi+2pin#

#sinx+1=0#, #=>#, #sinx=-1#, #=>#, #x=3/2pi+2pin#

#sinx-1=0#, #=>#, #sinx=1#, #=>#, #x=1/2pi+2pin#

Nghi N
Jul 16, 2017

#0; pi/4; pi/2; pi#

Explanation:

#f(x) = sin^2 3x - sin^2 x = (sin 3x - sin x)(sin 3x + sin x)#
Apply the 2 trig identities:
#sin a - sin b = 2cos((a + b)/2)sin ((a - b)/2)#
#sin a + sin b = 2sin ((a + b)/2)cos ((a - b)/2)#
In this case -->
f(x) = (2cos 2x.sin x)(2sin 2x.cos x) = 0
f(x) = (2sin 2x.cos 2x)(2sin x.cos x) = sin 4x.sin 2x = 0
Either factor must be zero.

a. sin 2x = 0 --> 2x = 0; #2x = pi#, and #2x = 2pi# -->
#x = 0; x = pi/2#; and #x = pi#
b. sin 4x = 0 --> 4x = 0; #4x = pi#; and #4x = 2pi #-->
#x = 0; x = pi/4#; and #x = pi/2#
Answers in interval #(0, 2pi)#:
#0; pi/4; pi/2; pi#
Check by calculator:
#x = pi/4# --> #sin x = 1/sqrt2# --> #sin^2 x = 1/2# -->
#sin 3x = sin ((3pi)/4) = sqrt2/2# --> #sin^2 ((3pi)/4) = 2/4 = 1/2#.
#sin^2 3x - sin^2 x = 1/2 - 1/2 = 0# Proved.

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