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#