Question

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

Answer 1

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.

Answer 2

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

Explanation:

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`