Solve $\cos(x) = \cos(x+a)$, if $\cos(x)\ge0$

Question

Solve $\cos(x) = \cos(x+a)$, if $\cos(x)\ge0$

This is how I did it:

For $\cos(x)\ge0, x\in[(4n-1)\frac\pi2,(4n+1)\frac\pi2]$

Also, $\cos(x+a)\ge0\implies x+a\in[(4m-1)\frac\pi2,(4m+1)\frac\pi2]$

$$\cos(x+a)=\cos(x)\\x+a=2n\pi\pm x$$
Case I: Taking the $+$ sign,
$a=2n\pi$
For these values of $a$, $x\in[(4n-1)\frac\pi2,(4n+1)\frac\pi2]$. This also satisfies the condition that $\cos(x+a)\ge0$
Case II Taking the $-$ sign,
$x=\frac{2n\pi+a}{2}$
How do I solve the rest of it, satisfying the conditions on $x$ and $a$ simultaneously? Should I replace $x$ with $\frac{2n\pi+a}{2}$ in the first two conditions or do something else?

Answer 1

Like Mathmo says, solve the resulting equation after expanding using the identity for cosine of the sum:

$$\cos(x)=\cos(x+a)\Rightarrow$$ $$\cos(x)=\cos(x)\cos(a)-\sin(x)\sin(a)\Rightarrow$$ $$\cos(x)\cdot(\cos(a)-1)=\sin(x)\sin(a)\Rightarrow$$ $$\frac{\sin(x)}{\cos(x)}=\frac{-1+\cos(a)}{\sin(a)}\Rightarrow$$ $$\tan(x)=\frac{-1+\cos(a)}{\sin(a)}\Rightarrow$$ $$x=\arctan\left(\frac{-1+\cos(a)}{\sin(a)}\right )$$

Answer 2

Given

$$ \cos(x+a) = \cos(x). $$

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Write this as

$$ \cos(x+a) - \cos(x) = 0, $$

thus

$$ \cos\Big( (x+a/2) + a/2 \Big) - \cos\Big( (x+a/2) - a/2 \Big) = 0, $$

which can be written as

$$ 2 \sin\Big( x+a/2 \Big) \sin\Big( a/2 \Big) = 0, $$

therefore

$$ a = 2 k \pi \vee a = 2k\pi - 2x. $$

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As

$$ \cos(x) \ge 0, $$

we have

$$ x \in \Big[2\ell\pi - \tfrac{1}{2}\pi,2\ell\pi + \tfrac{1}{2}\pi\Big], $$.

so

$$ 2x \in \Big[4\ell\pi - \pi,4\ell\pi + \pi\Big] $$

and

$$ 2x - 2k\pi \in \Big[2 (2\ell - k) \pi - \pi,2(2\ell-k)\pi + \pi\Big], $$

but

$$ \Big[2 (2\ell - k) \pi - \pi,2(2\ell-k)\pi + \pi\Big] = \Big[2 m - \pi, 2 m + \pi \Big] = \mathbb{R}, $$

so

$$ a \in \mathbb{R}. $$

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For a given $a \in \mathbb{R}$, we have

$$ a - 2\pi \Big\lfloor \frac{a+\pi}{2\pi} \Big\rfloor \in \Big[-\pi,+\pi\Big], $$

so

$$ - \tfrac{1}{2} a + \pi \Big\lfloor \frac{a+\pi}{2\pi} \Big\rfloor \in \Big[-\tfrac{1}{2}\pi,\tfrac{1}{2}\pi\Big], $$

whence

$$ 2\ell\pi - \tfrac{1}{2} a + \pi \Big\lfloor \frac{a+\pi}{2\pi} \Big\rfloor \in \Big[2\ell\pi-\tfrac{1}{2}\pi,2\ell\pi-\tfrac{1}{2}\pi\Big]. $$

We also have

$$ x = k\pi - \tfrac{1}{2} a \in \Big[2\ell\pi-\tfrac{1}{2}\pi,2\ell\pi-\tfrac{1}{2}\pi\Big], $$

therefore

$$ x = 2\ell\pi - \tfrac{1}{2} a + \pi \Big\lfloor \frac{a+\pi}{2\pi} \Big\rfloor \in \Big[2\ell\pi-\tfrac{1}{2}\pi,2\ell\pi-\tfrac{1}{2}\pi\Big], $$

and the solution for $x$.