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

##### 2 Answers

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

so:

so:

and:

(remembering that

There is another way:

(remembering that

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

and

#### Explanation:

The sine double-angle formula is

#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

#sin(2theta)=0#

Thinking about when the sine function is

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

Which can be generalized when

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

Note that

#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

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

Solving for

#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#