Question

How do you solve `sin^2x-4sinx-5=0` and find all general solutions?

Answer 1

`x = (3pi)/2 + 2n pi` `{n in Z}`

Explanation:

`sin^2x-4sinx-5 =0`

Let `phi -=sin x ->` `phi^2 -4phi -5 =0`

Factorise: `(phi+1)(phi-5) = 0`

Therefore `phi =` either `-1 or 5`
Since `phi = sinx ->` `-1<= phi <= 1` Therefore `phi` cannot `= 5`

Hence `phi =-1`

Since `phi = sin x`

`sin x =-1`
Therefore `x = arcsin(-1) = (3pi)/2` for `x in (0,2pi)`

Since the question asked for a genral solution:
`x = (3pi)/2 +2n pi` `{n in Z}`
Since the period of the `sin` function is `2pi`.

Answer 2

`x=(3pi)/(2)`

Explanation:

We have: `sin^(2)(x)-4sin(x)-5=0`

This trigonometric equation is in the form of a quadratic, so let's factorise it:

`=> sin^(2)(x)+sin(x)-5sin(x)-5=0`

`=> sin(x)(sin(x)+1)-5(sin(x)+1)=0`

`=> (sin(x)-5))(sin(x)+1)=0`

`=> sin(x)=5 =>` undefined

or

`=> sin(x)=-1 => x=(3pi)/(2)`

Therefore, the solution to the equation is `x=(3pi)/(2)`.