How do you solve #sin(2theta) + sin(4theta) = 0#?

You can solve it with the formula sum-to-product, that says:

#sinalpha+sinbeta=2sin((alpha+beta)/2)cos((alpha-beta)/2)#,

so:

#sin2theta+sin4theta=0rArr2sin((2theta+4theta)/2)cos((2theta-4theta)/2)=0rArr#

#sin3thetacos(-theta)=0#

so:

#sin3theta=0rArr3theta=kpirArrtheta=kpi/3#

and:

#cos(-theta)=0rArrcostheta=0rArrtheta=pi/2+kpi#.

(remembering that #cos(-alpha)=cosalpha#).

There is another way:

#sin2theta=-sin4thetarArrsin3theta=sin(-4theta)#

(remembering that #sin(-alpha)=-sinalpha#),

and than remembering that two sini are equal when the angles are equal or if the angles are supplementary,

#2theta=-4theta+2kpirArr6theta=2kpirArrtheta=kpi/3#

and

#2theta=pi-(-4theta)+2kpirArr2theta=pi+4theta+2kpirArr#

#-2theta=pi+2kpirArrtheta=-pi/2-kpi# that is formally identical to the second solution found before.

The sine double-angle formula is #sin(2alpha)=2sin(alpha)cos(alpha)#. Thus, #sin(4theta)=2sin(2theta)cos(2theta)#. The given equation is then equivalent to

#sin(2theta)+2sin(2theta)cos(2theta)=0#

Factoring yields

#sin(2theta)(1+2cos(2theta))=0#

Now, we can solve the two resulting equations whose product is #0#. The first gives

#sin(2theta)=0#

Thinking about when the sine function is #0#, we see that

#2theta=0,pi,2pi,3pi...#

Which can be generalized when #k# is an integer by saying

#2theta=kpi" "," "kinZZ#

Note that #kinZZ# is the symbolic way of representing that #k# is an integer. Thus

#color(blue)(theta=(kpi)/2" "," "kinZZ#

The other resultant equation from before was

#1+2cos(2theta)=0#

So

#cos(2theta)=-1/2#

This happens at

#2theta=(2pi)/3,(4pi)/3#

And all of these angles coterminal versions, which are located at integer multiples of #2pi# away. Thus

#2theta=(2pi)/3+2kpi,(4pi)/3+2kpi" "," "kinZZ#

Solving for #theta# yields

#color(blue)(theta=pi/3+kpi,(2pi)/3+kpi" "," "kinZZ#

We can combine all of our solutions into

#color(red)(theta=(kpi)/2,pi/3+kpi,(2pi)/3+kpi" "," "kinZZ#