How do I solve $\sin^2 x=\cos x$?

Question

I'm trying to solve a trigonometric equation, but I'm a bit stuck. The equation is this:

$\sin^2 x = \cos x$

So far what I've done looks like this:

$\sin^2 x - \cos x = 0$

$ (1 - \cos^2 x) - \cos x = 0$

$-\cos^2 x - \cos x + 1 = 0$

But from there I don't know how to factor it to get onwards to evaluating $x$ for separate cosine terms. Have I gone wrong somewhere, or am I simply not seeing the proper way to factor this?

Just solve it as you would solve any quadratic equation, just not in $x$, but in $\cos x$. — AnotherTest, Dec 23 '13 at 21:58
In other words, let $y=\cos x$. Solve for $y$. Then solve for $x$. — user7530, Dec 23 '13 at 21:59
@amWhy: I think you made a sign error. This quadratic has real roots. — Alex Kruckman, Dec 23 '13 at 22:09
@AlexKruckman, yep, $\sin^2x$ definitely intersects $\cos x$ in $(0, \pi/2)$. — Kaster, Dec 23 '13 at 22:10

Adi Dani · Accepted Answer · 2013-12-23T22:26:48.623

10

$$\sin^2 x = \cos x$$ $$1-\cos^2 x -\cos x= 0$$ $-1\leq\cos x=t\leq 1$ $$t^2+t-1=0$$ $$t_{1}=\frac{-1+\sqrt{5}}{2}\in[-1,1],t_{2}=\frac{-1-\sqrt{5}}{2}<-1,t_2\notin[-1,1]$$ $$x_1=\arccos t_1$$

edited Dec 23 '13 at 22:26

answered Dec 23 '13 at 22:13

Adi Dani

16,949

but $$t_{1}=\frac{-1-\sqrt{5}}{2}<1$$ – nadia-liza Dec 23 '13 at 22:15
2

@nadia-liza: You mean "$< -1$". – TonyK Dec 23 '13 at 22:21
1

@nadia-liza: $t_1$ is obviously not a solution, but $t_2=\arccos\dfrac1\phi$ however is. – Lucian Dec 23 '13 at 22:25
@nadia-liza. My answer is edited. – Adi Dani Dec 23 '13 at 22:28
Can anyone tell me why the right square bracket in the above expression $\interval{-1,1}$ prints more faintly than the left square bracket? I've noticed this throughout MSE. It may be a LaTex thing. – Guy Corrigall Dec 24 '13 at 00:17
I have absolutely no idea what's happening here... I don't understand this answer. – Threethumb Dec 24 '13 at 10:14
Actually, after trying it out on paper myself, it clicked. I get it now, thanks! – Threethumb Dec 24 '13 at 10:29

How do I solve $\sin^2 x=\cos x$?

1 Answers1