Question

How do you find the exact value of `2sin2theta+sintheta=0` in the interval `0<=theta<2pi`?

Answer 1

`theta={0,1.82,3.14,4.46}` (in radians up to `2dp`)

Explanation:

As `2sin2theta+sintheta=0`, we have

`2xx2sinthetacostheta+sintheta=0`

or `sintheta(4costheta+1)=0`

hence either `sintheta=0` i.e. `theta=0` or `pi` (in radians)

or `4costheta+1=0` i.e. `costheta=-1/4=cos1.82` or `cos(2pi-1.82)=cos4.46` (in radians up to `2dp`)

i.e. `theta={0,1.82,3.14,4.46}`