Question

How do you solve `\sin \frac { 2x + 1} { x } + \sin \frac { 2x + 1} { 3x } - 3\cos ^ { 2} \frac { 2x + 1} { 3x } = 0`?

Answer 1

`x=1/(3arcsin((-1+sqrt(10))/3)+6npi-2)` or `x=1/(3(2n+1)pi-3arcsin((-1+sqrt(10))/3)-2)` where `n in Z`.

Explanation:

Let `(2x+1)/(3x)=u`.

Then,

`sin(u)+sin(u)-3cos^2(u)=0`.

Using the trigonometric identity `sin^2(u)+cos^2(u)=1`,

`2sin(u)-3(1-sin^2(u))=0`,
`3sin^2(u)+2sin(u)-3=0`.

This is a quadratic equation in `sin(u)` it has roots given by the quadratic formulae of

`sin(u) = (-2pmsqrt(4+4*9))/(2*3)`,
`sin(u) = (-1pmsqrt(10))/(3)`.

Then, the general solution to `sin(u)=K` is `u=arcsin(K)+2npi` or `u = pi(1+2n) - arcsin(K)` where `n in Z`.

Notice that `sqrt(10)>3` so then `-1-sqrt(10)<-4/3`. So then the only solution set we need consider is `sin(u) = (-1+sqrt(10))/(3)`, as `abs(sin(x))<=1`.

Then, the solution set for `sin(u)` is

`u=arcsin((-1+sqrt(10))/3)+2npi` or `u=-arcsin((-1+sqrt(10))/3)+(2n+1)pi` where `n in Z`.

If `(2x+1)/(3x)=K` then,

`2x+1=3Kx`,
`x(2-3K)=-1`,
`x=1/(3K-2)`.

Then the solution set for `x` is

`x=1/(3arcsin((-1+sqrt(10))/3)+6npi-2)` or `x=1/(3(2n+1)pi-3arcsin((-1+sqrt(10))/3)-2)` where `n in Z`.

Answer 2

Let `a=(2x+1)/(3x)`

So the given equation becomes

`sin3a+sina-3cos^2a=0`

`=>3sina-4sin^3a+sina-3cos^2a=0`

`=>4sina-4sin^3a-3cos^2a=0`

`=>4sina(1-sin^2a)-3cos^2a=0`

`=>4sinaxxcos^2a-3cos^2a=0`

`=>cos^2a(4sina-3)=0`

When `cos^2a=0`

`=>cosa=0`

`=>a=(2n+1)pi/2" where "n in ZZ`

`=>(2x+1)/(3x)=(2n+1)pi/2" where "n in ZZ`

`=>x((6n+3)pi/2-2)=1`

`=>x=1/((6n+3)pi/2-2)" where " n in ZZ`

When `sina=3/4=sinalpha`,

where `alpha=sin^-1(3/4)`

`=>a=npi+(-1)^nalpha" where " n in ZZ`

`=>(2x+1)/(3x)=npi+(-1)^nalpha" where " n in ZZ`

`=>x=1/(3(npi+(-1)^nalpha)-2)" where " n in ZZ`