How do you solve #Sin 3 theta = sin theta #?

3 Answers

#theta=tan^(-1)1=pi/4 +k*2pi or (5pi)/4+m*2pi# ; #k,m in ZZ#.

Explanation:

#sin3theta=sintheta#

#therefore sin(2theta+theta)=sintheta#

#thereforesin2thetacostheta+cos2thetasintheta=sintheta#

#therefore2sinthetacosthetacostheta+(cos^2theta-sin^2theta)sintheta=sintheta#

#therefore2cos^2theta+cos^2theta-sin^2theta=1#

#therefore2cos^2theta-sin^2theta=1-cos^2theta=sin^2theta#

#therefore2cos^2theta=2sin^2theta#

#thereforecostheta=sintheta#

#therefore 1=(sintheta)/(costheta)=tantheta>0#

#therefore theta=tan^(-1)1=pi/4 +k*2pi or pi+pi/4+m*2pi# ; #k,m in ZZ#.

Jun 12, 2016

There are two sets of solutions:
#theta = pi n#
#theta = pi/4+pi/2 n#
(where #n# - any integer number)

Explanation:

Recall the derivation of a formula for #sin 3theta#:
#sin 3theta = sin(theta+2theta)#
#= sin theta * cos 2theta + sin 2theta * cos theta#
#= sin theta * (cos^2 theta-sin^2 theta) + 2sin theta*cos theta*cos theta#
#= sin theta * (1-2sin^2 theta) + 2sin theta * (1-sin^2 theta)#
#= 3sin theta - 4 sin^3 theta#

Applying this to our equation, we get
#3sin theta - 4 sin^3 theta = sin theta#
or
#2sin theta - 4sin^3 theta = 0#
or
#sin theta*(1-2sin^2 theta) = 0#

This equation has one set of solutions when #sin theta = 0#, that is
(Solution 1.1) #theta = 0+2pi n# and
Solution 1.2) #theta=pi+2pi n#,
which can be combined into
(Solution 1') #theta = pi n#, where #n# - any integer number.

Another set of solution is from
#1-2sin^2 theta = 0#
or
#sin theta = +-sqrt(2)/2#
With "#+#" sign the solutions are
(Solution 2.1) #theta = pi/4 +2pi n# and
(Solution 2.2) #theta = (3pi)/4+2pi n#
With "#-#" sign the solutions are
(Solution 3.1) #theta = -pi/4+2pi n# and
(Solution 3.2) #theta = -(3pi)/4 + 2pi n#
In both cases #n# is any integer number.

NOTE: Solutions 2.1, 2.2, 3.1 and 3.2 can be combined into one expression: #theta = pi/4+pi/2n#, where #n# - any integer number.

CHECK (we can ignore #2pi n# since #2pi# is a period for all participating functions)
Solution 1.1:
Left side equals
#sin (3*0) = 0#
Right side equals
#sin 0 = 0#
Solution 1.2:
Left side equals
#sin (3pi) =# [since #2pi# is a period] #= sin (3pi-2pi) = sin(pi) = 0#
Right side equals
#sin(pi) = 0#
Solution 2.1:
Left side equals
#sin (3*pi/4) = sin (pi-pi/4) = # [since #sin phi=sin(pi-phi)#] #= sin(pi/4) = sqrt(2)/2#
Right side equals
#sin(pi/4) = sqrt(2)/2#
Solution 2.2:
Left side equals
#sin (3*(3pi)/4) = sin ((9pi)/4)=sin((9pi)/4-2pi) = sin(pi/4) = sqrt(2)/2#
Right side equals
#sin ((3pi)/4) = sqrt(2)/2# (see 2.1 above)
Solution 3.1:
Left side equals
#sin (3*(-pi)/4) = #[since #sin(-phi)=-sin(phi)#] #= -sin ((3pi)/4) = - sqrt(2)/2#
Right side equals
#sin((-pi)/4) = #[since #sin(-phi)=-sin(phi)#] #=-sqrt(2)/2#
Solution 3.2:
Left side equals
#sin (3*(-3pi)/4) = sin ((-9pi)/4)=sin((-9pi)/4+2pi) = sin(-pi/4) = -sqrt(2)/2#
Right side equals
#sin ((-3pi)/4) = -sin((3pi)/4) = -sqrt(2)/2#

Jul 26, 2016

Either

#theta=(npi)/2+pi/4,"where "n in ZZ#

Or
#theta=kpi,"where "k in ZZ#

Explanation:

The given equation

#sin3theta=sintheta#

#=>sin3theta-sintheta=0#

#=>2cos((3theta+theta)/2)sin((3theta-theta)/2)=0#

#=>2cos(2theta)sintheta=0#

Either
#:.cos(2theta)=0#

#2theta=(2n+1)pi/2=npi+pi/2 #

#=>theta=(npi)/2+pi/4,"where "n in ZZ#

Or,

#sintheta=0#

#theta=kpi,"where "k in ZZ#