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Given the equation: $\sin^2{x}+\cos{x}=0$

How is it solved?

I think: $\sin^2{x}=1-\cos^2{x}$, but even if I get a quadratic equation with one function (cos), how can I solve it?

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  • $\begingroup$ Substitute $u = \cos x$. Solve the resulting quadratic, and then take the inverse cosine of the results. Discard results that don't lie between $-1$ and $1$. $\endgroup$
    – Tom
    May 12 '18 at 22:53
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    $\begingroup$ Apply it the same way: $-\cos^2 x + \cos x + 1=0$. $\cos x = {-1 \pm \sqrt{1+4} \over -2}$. $\endgroup$ May 12 '18 at 22:54
  • $\begingroup$ @ChristopherMarley and find the arccos? $\endgroup$
    – Dave
    May 12 '18 at 23:00
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You are on the right track. The equation $$\sin^2 x + \cos x = 0 $$ becomes $$-\cos^2 x + \cos x + 1 = 0 $$ with the substitution $\sin^2 x = 1 - \cos^2 x$. At this point, you may solve the quadratic equation in $\cos x$ to find $$\cos x = \frac{-1 \pm \sqrt{1 + 4}}{-2} = \frac{1 \mp\sqrt 5}{2},$$ that is, formally, $$\cos x = \begin{cases} \varphi \\ - \dfrac 1 \varphi \end{cases} $$ where $\varphi = 1.618033...$ is the golden ratio. However, since $\cos x \in [-1,+1]$, the option $\cos x = \varphi$ must be discarded, so that $$\cos x = - \frac 1 \varphi = \frac{1 - \sqrt 5}{2}. $$ One solution is $$x = \arccos \frac{1 - \sqrt 5}{2} \approx 2.237$$ in radians (in degrees, about $128.2^\circ$); however, since $\cos$ is an even function, $-x$ must be a solution too. Finally, $\cos$ is $2\pi$-periodic, therefore the other solutions may be found by adding a factor of $2 \pi n$ to $x$ and $-x$, with $n \in \mathbb Z$.

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  • $\begingroup$ Thank you very much for your answer! $\endgroup$
    – Dave
    May 12 '18 at 23:33
  • $\begingroup$ You're welcome! $\endgroup$
    – giobrach
    May 12 '18 at 23:34
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$$\sin^2{x}+\cos{x}=0$$

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

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

$$ \cos x = \frac {1-\sqrt 5 }{2}$$

$$x= \cos ^{-1} (\frac {1-\sqrt 5 }{2}) \approx 128.17 \text { degrees.}$$

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