Question

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

Answer 1

`x=0+2pik,`

`color(white)(x=)pi+2pik,`

`color(white)(x=)(2pi)/3+2pik,` and

`color(white)(x=)(4pi)/3+2pik`

Or, simplified:

`x=(2pi)/3*kquad` and `quadpik`.

Explanation:

`sin(2x)+sinx=0`

`2sinxcosx+sinx=0`

`sinx(2cosx+1)=0`

`sinx=0,qquadcosx=-1/2`

Here's a unit circle to remind us of where the sine and cosine values are:

This means that:

`x=0,pi,(2pi)/3,(4pi)/3`

Since these values are the same after any full `2pi` rotation, we write `+2pik` after every solution to represent that the answer will stay the same after any rotation added or subtracted:

`x=0+2pik,qquadpi+2pik,qquad(2pi)/3+2pik,qquad(4pi)/3+2pik`

Technically, our answers are complete, but we can go a little further.

We can see a pattern in some of the solutions: `0, (2pi)/3, (4pi)/3`

Since these are multiples of `(2pi)/3`, we can rewrite these three solutions as `(2pi)/3*k`. Now, our solutions are:

`x=(2pi)/3*k,qquadpi+2pik`

Since `0` is still a solution, there is another pattern: `0, pi, 2pi`

We can rewrite this again as multiples of `pi`, or `pik`. Now, our final-final solutions are:

`x=(2pi)/3*k,qquadpik`

These are the solutions. Hope this helped!

Answer 2

`x= 2npi + (2pi)/3 or 2npi -(2pi)/3 or npi`

Explanation:

put `sin(2x)=2sin(x)cos(x)`

equation will now be reduced to

`2sin(x)cos(x)+sin(x)=0`

taking `sin(x)` common we get

`sin(x)(2cos(x)+1)=0`

two possibilities are there

either `sin(x)=0` or `2cos(x)+1=0`

if `sin(x)=0` then `x=npi`

if `2cos(x) +1 = 0` then `x=2npi + (2pi)/3 or 2npi -(2pi)/3`

so solution to the equation is

`x= 2npi + (2pi)/3 or 2npi -(2pi)/3 or npi`