Find the $y$-coordinate given $\cosθ + \sin^2θ = -\dfrac5{16}$.

Question

I've been stuck on this trying to solve it on my own and with the textbook instructions but the lesson isn't in depth enough to help me solve this.

I am given the equation $\cos θ + \sin^2θ = -\dfrac{5}{16}$

I know $\sin^2θ + \cos^2θ = 1$, but my $\cosθ$ isn't squared. I've tried squaring everything but then I end up with $\sin^4θ$ and more confusion...

I got the question wrong so many times I can't try anymore but I got a window with a message saying:

In a right​ triangle, with legs a and b and hypotenuse​ $c$, the Pythagorean Theorem states that a squared plus b squared equals $c^2,\,$ $a^2+b^2=c^2$. Use the given equation with the Pythagorean Theorem to write an equation using only one of the trigonometric functions. Remember that $\cos(\pi) = -1$ and $\sin(3\pi/2) = -1.$

I've run out of different ways to attempt this. Where can I go from here?

Answer 1

It's $$\cos\theta+1-\cos^2\theta=-\frac{5}{16},$$ which gives $\cos\theta=\frac{7}{4},$ which is impossible, or $\cos\theta=-\frac{3}{4}$ and $$\theta=\pm\left(\pi-\arccos\frac{3}{4}\right)+2\pi k,$$ where $k\in\mathbb Z$.

Answer 2

Use $\sin^2 \theta = 1-\cos^2 \theta$ to write the equation as $$\cos^2 \theta - \cos \theta - \frac {21}{16}=0.$$ Let $x=\cos \theta$ and solve the quadratic equation in $x$.