Trigonometric equations with 2 functions

Question

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?

Answer 1

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 $n \in \mathbb Z$.

Answer 2

$$\sin^2{x}+\cos{x}=0$$

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

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

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

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