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I have this equation $\cos2x +5 \cos x + 3=0$. To solve it I rewrite $\cos2x$ to $2 \cos^{2} x- 1$ and set $\cos = t$.

I get the following equation $2t^2 - 1 +5t +3 = 0$ with that and then divide the equation with two $t^2 +\frac{5}{2} t +1 = 0$. I solve this equation and get two $t$, $t_1 = -2$ and $t_2 = - \frac {1}{2}$. $t_2$ is the valid because $t$ can't be larger than 1.

From here on I don't know how to use $t$ to solve this equation $\cos2x +5 \cos x + 3=0$.

Can anyone explain what to do next and how to solve this equation?

Thanks!!

S4M1R
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3 Answers3

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It is not necessary to use substitutions when what you're working with isn't going to be too long:

$$ \cos 2x +5 \cos x + 3 = 0\\ \implies 2\cos^2 x - 1 + 5\cos x + 3 = 0\\ \implies 2\cos^2 x + 5\cos x + 2 = 0\\ \implies \cos^2 x + \frac{5}{2}\cos x + 1 = 0\\ \implies \cos^2 x + \left(2 + \frac{1}{2}\right)\cos x + 2\cdot \frac{1}{2} = 0\\ \implies (\cos x + 2)(\cos x + \frac{1}{2}) = 0 $$

Now, $\cos x \neq -2$ since $\cos x \in \left[-1,1\right]$. So we can reject that.

$$\therefore \space\cos x = - \frac{1}{2} \\ \implies \cos x = -\sin{\frac{\pi}{6}} = \cos\left(\frac{\pi}{2} + \frac{\pi}{6}\right) = \cos \left(\frac{2\pi}{3}\right)\\ \implies x = 2n\pi \pm \frac{2\pi}{3} , \space n\in \mathbb Z\\ \implies x = 2\pi \left(n \pm \frac{2}{3}\right), \space n\in \mathbb Z $$

Trivial yet Important Facts used:
  • $x^2 + \left(\alpha + \beta\right)x + \alpha\beta = 0 \iff (x+\alpha)(x+\beta) = 0$
  • $\cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta$
  • $\cos x = \cos \theta \implies x = 2n\pi \pm \theta$
Nick
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  • It should be $\cos x\ne -2$. – egreg Sep 29 '14 at 09:26
  • @egreg: Thanks, I've corrected it. But you didn't notice my bigger blunder of saying $-\cos \theta = \cos(\frac{\pi}{2} + \theta)$. That only works if $\theta = \frac{1}{2}$. I corrected that too. Please let me know if there are any more mistakes. I'm such a careless goose. – Nick Sep 29 '14 at 09:41
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Hint

You properly did the work. But, for sure, the solution $t=-2$ must be discarded (except if you work with complex numbers). So, the only solution is $t = - \frac {1}{2}$ from which $\cos(x)= - \frac {1}{2}$ and then $x=???$

I am sure that you can take from here.

Claude Leibovici
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1

$$\cos \color{green}{x} = -\frac{1}{2} \implies \cos x= \cos\left({\frac{\pi}{2}}^c +30^0 \right)$$

So, the general solution to the above equation is:

$$ x= 2n\pi +(\pi+\pi /6)\\ \text{or} \\ x=2n\pi -(\pi +\pi /6)$$

where $n = 0,1,2,3\dots$

Nick
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Jasser
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  • To make this answer unique, I suggest you elaborate on why $ \cos x= \cos\left({\frac{\pi}{2}}^c +30^0 \right)$ – Nick Sep 29 '14 at 10:21