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

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 $\cos x$. Dec 23, 2013 at 21:58
• In other words, let $y=\cos x$. Solve for $y$. Then solve for $x$. Dec 23, 2013 at 21:59
• @amWhy: I think you made a sign error. This quadratic has real roots. Dec 23, 2013 at 22:09
• @AlexKruckman, yep, $\sin^2x$ definitely intersects $\cos x$ in $(0, \pi/2)$. Dec 23, 2013 at 22:10

$$\sin^2 x = \cos x$$ $$1-\cos^2 x -\cos x= 0$$ $-1\leq\cos x=t\leq 1$ $$t^2+t-1=0$$ $$t_{1}=\frac{-1+\sqrt{5}}{2}\in[-1,1],t_{2}=\frac{-1-\sqrt{5}}{2}<-1,t_2\notin[-1,1]$$ $$x_1=\arccos t_1$$
• but $$t_{1}=\frac{-1-\sqrt{5}}{2}<1$$ Dec 23, 2013 at 22:15
• @nadia-liza: You mean "$< -1$". Dec 23, 2013 at 22:21
• @nadia-liza: $t_1$ is obviously not a solution, but $t_2=\arccos\dfrac1\phi$ however is. Dec 23, 2013 at 22:25
• 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. Dec 24, 2013 at 0:17