Question

How do you determine all solutions for the equation `2sin^2theta=1-sintheta` in the domain `0^circ<=theta<360^circ`?

Answer 1

The solutions are `S={270º,30º,150º}`

Explanation:

Let's rewrite the equation

`2sin^2theta+sin theta-1=0`

This s a quadratic equation, `ax^2+bc+c=0`

We calculate the discriminant,

`Delta=b^2-4ac=1-4*2*-1=9`

`Delta >0`, so we have 2 real roots

`x=(-b+-sqrt(Delta))/(2a)`

`sintheta=(-1+-sqrt9)/4=(-1+-3)/4`

`sintheta=-1` or `sintheta=1/2`

We are looking for the solutions, `theta in [0, 360[`

The solutions are `(270º, 30º,150º)`

Answer 2

`theta=pi/6`, `theta=5/6pi` and `theta=3/2pi`
for `0< theta <2pi`

Explanation:

rewrite the equation in an equivalent form

`2sin^2theta+sintheta-1=0`

this is a second order equation in the unknown `sintheta`

We can solve it, obtaining

`sintheta=(-1+-root2(1+8))/4=(-1+-3)/4`.
There are two possible solutions

`sintheta_1=-1` that means `theta_1=3/2pi+ 2 kpi`

or
`sintheta_2=1/2` that means `theta_2=pi/6+2kpi` or `theta_2=pi-pi/6+2kpi`

In total in `0< theta<2pi` we have three solutions
`theta=pi/6`, `theta=5/6pi` and `theta=3/2pi`