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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?

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    $\begingroup$ Reduce to a quadratic equation for $\cos\theta$. $\endgroup$ – Lord Shark the Unknown Sep 24 '17 at 17:51
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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 $k\in\mathbb Z$.

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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 $x$.

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  • $\begingroup$ Can you elaborate on how you used cos^2θ=1−sin^2θ to turn the equation into cos^2θ−cosθ−21/16=0 ? $\endgroup$ – Jason_Silva Sep 24 '17 at 18:53
  • $\begingroup$ You rearrange the identity $\cos^2 \theta + \sin^2 \theta =1$ to get that $\sin^2 \theta = 1- \cos^2 \theta/$. Plug this into the equation you're trying to solve, you get $\cos \theta + 1- \cos^2 \theta = -5/16$. Then bring everything to one side and group like terms to get $\cos^2 \theta - \cos - 21/16 =0$. Using the substitution $x=\cos \theta$ we get $x^2-x- \frac {21}{16}=0$ which has solutions $x=-3/4$ and $x=7/4$. So either $\cos \theta = -3/4$ or $\cos \theta = 7/4$. Note however that $-1 \leq \cos \theta \leq 1$ for any $\theta$ so only the first solution is possible. $\endgroup$ – Alex Sep 24 '17 at 18:56
  • $\begingroup$ The first time you wrote cos^2θ as positive and this time around you wrote it was negative. I'm not sure which one I want to use. $\endgroup$ – Jason_Silva Sep 24 '17 at 19:38
  • $\begingroup$ @Jason_Silva I changed it to positive so that the leading coefficient is 1, it doesn't matter which one you choose as you can multiply the equation by $-1$ to go between them. $\endgroup$ – Alex Sep 24 '17 at 20:00

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