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I'm trying to solve $\cos x=-\cos2x$, $-\pi<x\leq\pi$, WITHOUT USING the double angle formulae, here's my current working: $$\begin{align*}\cos x&=-\cos2x\\ \cos (x\pm2k\pi)&=\cos (2x\pm(2k+1)\pi)\\ \implies x\pm2k\pi&=2x\pm(2k+1)\pi\\ x&=2x\pm\pi\\ x&=\pm\pi\\ \therefore x&=\pi\text{ , in the given range of $x$}\end{align*}$$

Graphically it's obvious I've missed two other solutions, $x=\pm\frac{\pi}{3}$, however I do not know why because I have not, to my knowledge, carried out any illegal operations such as dividing by zero.

Could someone point out my mistake please and give an edited solution? Thanks!

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    $\begingroup$ $k$ on the left and right sides should not be the same. $\endgroup$ – Alex Silva Dec 12 '18 at 14:50
  • $\begingroup$ @AlexSilva I graphed them on Desmos for integer values of $k$ and they always lined up with the original graphs? Or am I missing something? $\endgroup$ – adam Dec 12 '18 at 15:02
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    $\begingroup$ Your implication is wrong. $\endgroup$ – Yves Daoust Dec 12 '18 at 15:05
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Use that

$$\cos x=-\cos2x \iff \cos x=\cos(\pi-2x)$$

and that

$$\cos \alpha = \cos \theta \iff \alpha = \theta +2k\pi \, \lor \, \alpha = -\theta +2k\pi \quad k\in \mathbb{Z}$$

Refer also to the related

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  • $\begingroup$ Ah okay, so from what I gathered (being still in highschool I'm not completely familiar with the if and only if implication so I've probably misinterpreted this answer), I came up with $\pm x+2k\pi = k\pi \pm 2x \implies k\pi = \pm x,\pm 3x$, which works but obviously only for odd-values of k, however k exists as an integer (and likewise works as an integer for my first statement in this comment)? $\endgroup$ – adam Dec 12 '18 at 15:17
  • $\begingroup$ @AdamBromiley The simbol $\lor$ stays for logical "or". $\endgroup$ – user Dec 12 '18 at 15:21
  • $\begingroup$ @AdamBromiley Then we have $$x=\pi-2x+2k\pi$$ or $$x=-(\pi-2x)+2k\pi$$ From here we can find all the solutions. Refer also to the given example. Do not hesitate to ask for any other clarification. Bye $\endgroup$ – user Dec 12 '18 at 15:23
  • $\begingroup$ Ah okay I get that now, apart from the one implication on the first line of your answer, why is there not a $k$ in $\cos(\pi -2x)$? Surely it is true for any odd multiples of pi? $\endgroup$ – adam Dec 12 '18 at 15:26
  • $\begingroup$ @AdamBromiley The first line is always true, we use that to transform the given equation in an other equivalent in order to eliminate the minus sign in front. Then we can use the second identity to find the solutions. $\endgroup$ – user Dec 12 '18 at 15:31
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$$\cos(x+\pi)=\cos(2x)\iff x+\pi=2k\pi\pm2x.$$

Hence

$$x=(2k+1)\pi\lor x=(2k+1)\frac\pi3.$$

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