How can I solve $\cos^2 x + \sin x +1 = 0$?

The solution set of the equation $$\cos^2 x + \sin x +1 = 0$$ is? I haven't studied trigonometry, I'm kinda lost on this issue ...

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Hint: $\sin^2x + \cos^2x = 1$. –  Cameron Williams Aug 17 '13 at 17:19
This raised a MathJax issue I hadn't seen before: the difference in appearance between $\displaystyle\cos ² x$ and $\displaystyle\cos^2 x$. (I changed it to the latter.) –  Michael Hardy Aug 17 '13 at 17:40
I fixed some grammar in your post, please see that the correction is correct. –  Jeel Shah Aug 18 '13 at 0:28

Using that $\sin^2x+\cos^2x=1$; obtain a quadratic in $\sin x$.

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Notice that since $-1 \le \cos x \le 1$ and $-1 \le \sin x \le 1$ we can see that $\cos^2 x \ge 0$ and $\sin x + 1 \ge 0$. Given $a,b \ge 0$ the only way $a+b = 0$ is if $a=0$ and $b=0$. Hence, we need both $\cos^2 x = 0$ and $\sin x + 1 = 0$. In other words, we need $\cos x = 0$ and $\sin x = -1$.

A quick look at a sine and cosine graph will show you that $x = 270^{\circ}$ is one possibility. In fact, its the only possibility between $0^{\circ}$ and $360^{\circ}$. Since the sine and cosine graphs repeat every $360^{\circ}$ we have $$(270 + 360n)^{\circ}$$ as possible solutions, where $n$ is any, possibly negative, whole number.

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+1 very nice a lot to learn from this answer –  Ovi Aug 17 '13 at 17:43
@Ovi Thank you, I appreciate your comment. –  Fly by Night Aug 17 '13 at 17:46
And this is a second post, altogether, after abandoning a now-deleted answer and with it, ridding yourself of your negative vote. –  amWhy Aug 17 '13 at 17:54
@amWhy It's a totally different method, and is totally unrelated to the other post. Why carry a negative vote from a completely unrelated post? The downvote was totally unwarranted anyway. A downvote is for an unuseful reply. A sign error does not make a post useless. –  Fly by Night Aug 17 '13 at 18:20
Great perspective man ! :) –  Simar Aug 18 '13 at 4:16

We can use the identity $\sin^2 x + \cos^2 x = 1$.

\begin{align} \cos^2 x + \sin x + 1 & = 0 \\ \\ (1 - \sin^2 x) + \sin x + 1 & = 0 \\ \\ \sin^2 x - \sin x - 2 & = 0\end{align}

Now, you have a quadratic in $\sin x$. For example, set $t = \sin x$, and you'll have a quadratic in $t$, find the roots, then find the corresponding solutions to each root.

Putting $\sin x = t,$ we obtain $$t^2 - t - 2 = (t-2)(t+1) = 0$$

Roots: $t = 2,\;\;t = -1$.

Since $t = \sin x$, that means that

• $t = \sin x = 2$ (impossible, since $-1 \leq \sin x \leq 1, \;\forall x \in \mathbb R$), or
• $t = \sin x = -1 \implies x = 3\pi/2 + 2k\pi$, where $k$ is any integer.
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Your $+1$ on the LHS became a $-1$ on the RHS. –  Fly by Night Aug 17 '13 at 17:26
Shouldn't it be $x = 2k\pi-\pi/2$? –  PepperSausage Aug 17 '13 at 17:44
@MathPhysUG That's still the same set of numbers. Replace your $k$ by $k+1$ to get amWhy's solution. –  Fly by Night Aug 17 '13 at 17:48
It was different before he changed it. –  PepperSausage Aug 17 '13 at 17:48
@MathPhysUG Fair point. It is a bit like the shifting sands. I think this is the fourth version now. –  Fly by Night Aug 17 '13 at 17:50

This is a, so-called, quadratic in disguise.

Use a familiar trigonometric identity linking $\cos^2 x$ and $\sin^2 x$ to put the equation in terms of $\sin x$. Then substitute $s=\sin x$. You'll notice that you have a quadratic equation in $s$. Just solve for $s$, then put your solutions back in terms of $x$ and solve for $x$.

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If you eliminate the trig functions you'll not only get something easy to solve but you'll also get some geometrical insight into the problem. $v=\sin\theta$ and $u=\cos\theta$ are the sides of a right triangle with hypotenuse $1$ and angle $\theta$. By Pythagoras we have $u^2+v^2=1$. The original equation is just $u^2+v+1=0$. You now have an easy quadratic equation to solve. But you can also see you're looking for where a certain parabola ($y = -x^2-1$) meets the unit circle $x^2+y^2=1$.

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