How do you solve #sin3x=cos3x#?

3 Answers
Jul 19, 2015

Use #tan 3x = (sin 3x) / (cos 3x) = 1# to find:

#x = pi/12 + (n pi) / 3#

Explanation:

Let #t = 3x#

If #sin t = cos t# then #tan t = sin t / cos t = 1#

So #t = arctan 1 + n pi = pi/4+n pi# for any #n in ZZ#

So #x = t/3 = (pi/4 + n pi) / 3 = pi/12 + (n pi) / 3#

Jul 19, 2015

Solve sin 3x = cos 3x

Answer: #x = pi/12 + Kpi/3#

Explanation:

Use the complementary arcs relationship:# cos x = sin (pi/2 - x)#

#sin 3x = sin (pi/2 - 3x)#
a. #3x = pi/2 - 3x# + 2Kpi -> #6x = pi/2 +2Kpi -> #
#x = pi/12 + Kpi/3#

Within interval# (0,2pi)# there are 6 answers: # pi/12; (5pi)/12; (9pi)/12; (13pi)/12; (17pi)/12; and (21pi)/12.#

b. #3x = pi - (pi/2 - 3x) = pi/2 + 3x.# This equation is undefined.

Check
#x = pi/12 --> sin 3x = sin pi/4 = sqrt2/2#
#x = pi/12 --> cos 3x = cos pi/4 = sqrt2/2#
Therefore sin 3x = cos 3x:
You may check other answers.

Jul 22, 2015

#x={(pi/12+(2pik)/3),(" "color(black)and),(-pi/4 + (2pik)/3):}#

#kinZZ#

Explanation:

Here's another method which has its own uses.

First, send every thing to one side

#=>sin(3x)-cos(3x)=0#

Next, express #sin3x-cos3x# as #Rcos(3x+lambda)#

#R# is a positive real and #lambda# is an angle

#=>sin(3x)-cos(3x)= Rcos(3x+lambda)#

#=>-cos(3x)+sin(3x)= Rcos(3x)coslambda-Rsin(3x)sinlambda#

Equate the coefficients of #cosx# and #sinx# on both sides

#=>" "Rcoslambda=-1 " "...color(red)((1))#
#" "-Rsinlambda=1 " "...color(red)((2))#

#color(red)(((2))/((1)))=>-(-Rsinlambda)/(Rcoslambda)=1/(-1)#

#=>tanlambda=1=>lambda=pi/4#

#color(red)((1)^2) + color(red)((2)^2) => (Rcoslambda)^2+(-Rsinlambda)^2=(-1)^2+(1)^2#

#=>R^2(cos^2lambda+sin^2lambda)=2#

#=>R^2(1)=2=>R=sqrt(2)#

So, #sin(3x)-cos(3x)=sqrt(2)cos(3x+pi/4)=0#

#=>cos(3x+pi/4)=0#

#=>3x+pi/4=+-pi/2+2pik#

Where #kinZZ#

Make #x# the subject

#=>x=+-pi/6-pi/12+2pik#

So we two sets of solutions :

#color(blue)(x={(pi/12+(2pik)/3),(" "color(black)and),(-pi/4 + (2pik)/3):})#

When #k=0=>x=pi/12+(2pi(0))/3=pi/12#

and #x=-pi/4+(2pi(0))/3=-pi/4#

When #k=1=>x=pi/12+(2pi)/3=(9pi)/12=(3pi)/4#
and #x=-pi/4+(2pi)/3=(5pi)/12#