Question

How do you solve `2sin^2x=5sinx+3`?

Answer 1

The solutions are `x=(7pi)/6+2pik,quad(11pi)/6+2pik`.

Explanation:

Treat it like a quadratic equation; you can make a substitution to make it easier to see:

`2sin^2x=5sinx+3`

Let `u=sinx`:

`2u^2=5u+3`

`2u^2-5u=3`

`2u^2-5u-3=0`

`2u^2-6u+u-3=0`

`color(red)(2u)(u-3)+color(blue)1(u-3)=0`

`(color(red)(2u)+color(blue)1)(u-3)=0`

`u=-1/2,3`

Now put back in `sinx` for `u`:

`sinx=-1/2, color(red)cancelcolor(black)(sinx=3)`
`" "color(white)*color(red)uarr`
This solution isn't possible, since `sin` has a range of `[-1,1]`.

To solve for `x` in the equation that's left, let's take a look at the unit circle to remind us of some `sin` values:

We can see that `sinx=-1/2` when `x` is `(7pi)/6` or `(11pi)/6`, so those are our solutions, but we are not completely done.

We need to show that these value will stay the same after any full rotation. (For instance, `x` can also be `-pi/6` because that is the same as rotating `(11pi)/6`.)

The way to represent this is by adding `2pik` to both solutions. `k` means "any integer." The `2pi` simply means a full rotation. So all together, it shows that the answer will be the same after any full rotation.

Here are the final results:

`x=(7pi)/6+2pik,quad(11pi)/6+2pik`

Hope this helped!