# Inverse Trig Equation

I was wondering if someone can give me a hint towards solving the following:

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

It feels like a quadratic equation kind of problem, but I'm supposed to use the inverse trig functions. What am I missing here? I feel it's something painfully obvious.

Thanks!

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First, you find the values of $\sin\theta$ that solve the equation, using the quadratic formula or some other method. Then you use inverse trigonometric functions to find the values of $\theta$.

Setting $x=\sin(\theta)$, we have to solve $$x^2 -x -1 =0.$$ Since $$x^2-x-1 = \left(x - \frac{1+\sqrt{5}}{2}\right)\left(x - \frac{1-\sqrt{5}}{2}\right),$$ then the two solutions are $$x = \frac{1+\sqrt{5}}{2},\quad\text{and}\quad \frac{1-\sqrt{5}}{2}.$$

Now, $1+\sqrt{5} \gt 1+2 = 3$, so $\sin(\theta) = \frac{1+\sqrt{5}}{2}$ has no solutions. Thus, the only possibility is for $\sin(\theta) = \frac{1-\sqrt{5}}{2}$.

That means that the principal value of $\theta$ that solves the equation is $$\theta = \arcsin\left(\frac{1-\sqrt{5}}{2}\right).$$ And of course, you also need to consider other values of $\theta$ outside of $[-\pi/2,\pi/2]$ that may yield the same value of $\theta$. There should be at least one other on $[-\pi,\pi]$, and then you can also add multiples of $2\pi$.

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Substitute $x = \sin(\theta)$ and solve the quadratic equation for $x$. Then use $\sin^{-1}(x)$ to solve for $\theta$.

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