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What is the general solution to this trig equation?

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

Thanks.

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  • $\begingroup$ What is going on? The question is massively upvoted (although it has zero context), as are some answers (even some that are wrong), for no valid reason that I can fathom... $\endgroup$
    – Did
    Jan 8 '17 at 17:27
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To make it easier to visualise and solve, you can substitute $u=\cos{x}$. $$2\cos^2{x}-\cos{x}=0$$ $$2u^2-u=0$$ From here, you can either use the quadratic formula or factorise. I chose to factorise: $$u(2u-1)=0$$ Hence we have the solutions $u=0,\frac{1}{2}$. Now, we solve the equations resulting from our substitution $u=\cos{x}$: $$\cos{x}=0 \tag{1}$$ $$\cos{x}=\frac{1}{2} \tag{2}$$

For equation $(1)$, we obtain:` $$x=\frac{\pi}{2}+k\pi$$ For equation $(2)$, we obtain: $$x=2k\pi+\frac{\pi}{3}$$ And $$x=2k\pi-\frac{\pi}{3}$$ Where $k \in \mathbb{Z}$.

Hence, our general solution is: $$x=\begin{cases} \frac{\pi}{2}+k\pi \\ 2k\pi+\frac{\pi}{3} \\ 2k\pi-\frac{\pi}{3} \end{cases}$$

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    $\begingroup$ The key thing for OP to understand is that one can't just divide thru by cos(x) as this would lose the first solution cos(x)=0 . The u substitution is brilliant as it helps prep students for the calculus to come. $\endgroup$ Jan 3 '17 at 14:31
  • $\begingroup$ Haha, you have been quite active now haven't you? $\endgroup$ Jan 10 '17 at 23:41
  • $\begingroup$ I must applaud you for making it to the first page : math.stackexchange.com/users $\endgroup$ Jan 10 '17 at 23:48
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hint: $2\cos^2 x -\cos x = \cos x(2\cos x - 1)=0$

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Notice that $$2 \cos^2 x-\cos x=0\implies 2\cos^2x=\cos x$$

So either $\cos x=0$ or $\cos x=\frac{1}{2}$.

Now find general solution

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$$2\cos^2(x) - \cos(x) = 0 \implies \cos(x)(2\cos(x) - 1) = 0$$

For what $x$ is $\cos(x) = 0, \frac{1}{2}$? There are only 2 solutions that form all distinct solutions; all others are $2k\pi{}$ away from them

Have a look

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    $\begingroup$ not correct - the solutions for cos(x) = 0 are more common than that $\endgroup$
    – Alnitak
    Jan 3 '17 at 14:46
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    $\begingroup$ There are * three* solutions that form all distinct solutions, namely $2n\pi \pm \frac{\pi}{3}$ and the solutions you neglect, $\pi n + \frac{\pi}{2}$ $\endgroup$ Jan 3 '17 at 19:05
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$2\cos^2 x-\cos x=0$

$\implies \cos x(2\cos x-1)=0$

Either

$\cos x=0$

$\implies x=\dfrac{\pi}{2}\pm 2n\pi$

Or

$2\cos x-1=0$

$\implies \cos x=\dfrac{1}{2}$

$\implies x=\dfrac{\pi}{3}\pm 2n\pi$

where $n \in Z$

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    $\begingroup$ Your missing some solutions. $\endgroup$ Jan 3 '17 at 6:03
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    $\begingroup$ also $-{\pi}/3+...$ $\endgroup$
    – Widawensen
    Jan 3 '17 at 12:48
  • $\begingroup$ You don't need to write $\pm$... Just let $n$ be negative as well. Moreover, you need to include $2n\pi - \frac{\pi}3$ $\endgroup$ Jan 3 '17 at 19:04