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How do you solve the following equation over an unrestricted domain; $$\left| \cos { \left( 2x \right) } \right| = \frac { 1 }{ 2 } $$

I can solve half of it;

$$\cos { \left( 2x \right) } =\frac { 1 }{ 2 } \\ x=\pi n\pm \frac { \pi }{ 6 } $$

but how do you do the other half $$\cos { \left( 2x \right) } =-\frac { 1 }{ 2 } $$

user222974
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3 Answers3

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We have $\cos(2x)=\pm\dfrac12$

$\iff\cos4x=2\cos^22x-1=\cdots=-\dfrac12=-\cos\dfrac\pi3=\cos\left(\pi-\dfrac\pi3\right)$

$$\implies4x=2m\pi\pm\dfrac{2\pi}3=\dfrac{2\pi}3(3m\pm1)$$ where $m$ is any integer

lab bhattacharjee
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First, draw the graph for $\cos \theta$ in the domain $[0,\ 2\pi]$. Then find the values of $\theta$ for which $\cos\theta = -\frac12$, and then replace $\theta$ with $2x$ and solve for $x$. To extend your answer to take into account an unrestricted domain, observe that for any value of $\theta$, adding $2\pi$ to $\theta$ doesn't change the value of $\cos \theta$. By the way, your answer to the first half is incomplete, it should be $x = \frac{\pi}{6} + n\pi$, $\frac{5\pi}{6} + n\pi$, where $n \in \mathbb Z$.

Martin Sleziak
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Vizuna
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  • Note that user222974's answer $$x = \pm \frac{\pi}{6} + n\pi$$ for the equation $$\cos(2x) = \frac{1}{2}$$ is equivalent to your answer $$x = \frac{\pi}{6} + n\pi \qquad \text{or} \qquad x = \frac{5\pi}{6} + n\pi$$ since $$\frac{5\pi}{6} = -\frac{\pi}{6} + \pi$$ – N. F. Taussig Mar 14 '15 at 13:06
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you have two equations $$\cos 2x = \frac 12 \to 2x = \pm \pi/3 + 2k\pi \\ \cos 2x = -\frac 12 \to 2x = \pm 2\pi/3 + 2k\pi$$ that is $$x = \pm \pi/6 + k\pi, \pm \pi/3 + k\pi \text{ where $k$ is any integer.} $$

abel
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