Can you solve 9sin^2theta - 6sintheta = 1 over [0 degree, 360 degrees) ?

This is just a quadratic equation, which we can easily solve:

Then:

Using quadratic formula:

For:

and

Approximate values:

To solve the equation $9\sin^2(\theta) - 6\sin(\theta) = 1$ over the interval $[0^\circ, 360^\circ)$, we can first rewrite it as a quadratic equation in terms of $\sin(\theta)$:

$9\sin^2(\theta) - 6\sin(\theta) - 1 = 0$

Now, let $u = \sin(\theta)$, then the equation becomes:

$9u^2 - 6u - 1 = 0$

This is a quadratic equation, which we can solve using the quadratic formula:

$u = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$

where $a = 9$, $b = -6$, and $c = -1$.

Plugging in the values:

$u = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4(9)(-1)}}}}{{2(9)}}$

$u = \frac{{6 \pm \sqrt{{36 + 36}}}}{{18}}$

$u = \frac{{6 \pm \sqrt{72}}}{{18}}$

$u = \frac{{6 \pm 6\sqrt{2}}}{{18}}$

$u = \frac{{1 \pm \sqrt{2}}}{{3}}$

Now, we have two possible solutions for $u$:

1. $u = \frac{{1 + \sqrt{2}}}{{3}}$
2. $u = \frac{{1 - \sqrt{2}}}{{3}}$

Since $\sin(\theta) = u$, we need to find the corresponding values of $\theta$ within the given interval $[0^\circ, 360^\circ)$.

For $u = \frac{{1 + \sqrt{2}}}{{3}}$: $\theta = \sin^{-1}\left(\frac{{1 + \sqrt{2}}}{{3}}\right)$

For $u = \frac{{1 - \sqrt{2}}}{{3}}$: $\theta = \sin^{-1}\left(\frac{{1 - \sqrt{2}}}{{3}}\right)$

Once we have obtained the values of $\theta$, we need to ensure they fall within the interval $[0^\circ, 360^\circ)$.