Question

How do you solve `tantheta = sqrt(2)sintheta`?

Answer 1

`theta = (pm k pi uu pm pi/4 + 2kpi)` for `k in ZZ`

Explanation:

`tan(theta)=sqrt(2)sin(theta)` or

`tan(theta)(1-sqrt(2)cos(theta))=0`

`{(tan(theta)=0->theta=pm k pi),(cos(theta)=1/sqrt(2)->theta = pm pi/4 + 2kpi):}`

`theta = (pm k pi uu pm pi/4 + 2kpi)` for `k in ZZ`

Answer 2

`theta={pi/4, (7pi)/4}`

Explanation:

Here's an alternative approach. Note that `tantheta = sin theta/costheta`.

`sintheta/costheta = sqrt(2)sintheta`

Multiply both sides by `costheta`.

`sintheta/costheta * costheta = sqrt(2)sintheta*costheta`

Now recognize that `sin2theta = 2sinthetacostheta`.

`sintheta = sqrt(2)sinthetacostheta`

Multiply the right side by `2/2`
`sintheta = 2sqrt(1/2)sinthetacostheta`

`sintheta = sqrt(1/2)sin2theta`

`sintheta/(2sinthetacostheta) = 1/sqrt(2)`

`1/(2costheta) = 1/sqrt(2)`

`sqrt(2) = 2costheta`

`sqrt(2)/2 = costheta`

This is the rationalized form of `1/sqrt(2)`.

`1/sqrt(2) = costheta`

This has solutions `theta = pi/4, (7pi)/4`.

Hopefully this helps!