3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Just solve it as you would solve any quadratic equation, just not in $x$, but in $\cos x$. $\endgroup$ – AnotherTest Dec 23 '13 at 21:58
  • 2
    $\begingroup$ In other words, let $y=\cos x$. Solve for $y$. Then solve for $x$. $\endgroup$ – user7530 Dec 23 '13 at 21:59
  • 1
    $\begingroup$ @amWhy: I think you made a sign error. This quadratic has real roots. $\endgroup$ – Alex Kruckman Dec 23 '13 at 22:09
  • $\begingroup$ @AlexKruckman, yep, $\sin^2x$ definitely intersects $\cos x$ in $(0, \pi/2)$. $\endgroup$ – Kaster Dec 23 '13 at 22:10
9
$\begingroup$

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

$\endgroup$
  • $\begingroup$ but $$t_{1}=\frac{-1-\sqrt{5}}{2}<1$$ $\endgroup$ – nadia-liza Dec 23 '13 at 22:15
  • 2
    $\begingroup$ @nadia-liza: You mean "$< -1$". $\endgroup$ – TonyK Dec 23 '13 at 22:21
  • 1
    $\begingroup$ @nadia-liza: $t_1$ is obviously not a solution, but $t_2=\arccos\dfrac1\phi$ however is. $\endgroup$ – Lucian Dec 23 '13 at 22:25
  • $\begingroup$ @nadia-liza. My answer is edited. $\endgroup$ – Adi Dani Dec 23 '13 at 22:28
  • $\begingroup$ 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. $\endgroup$ – Guy Corrigall Dec 24 '13 at 0:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.