$f(x) = A_1 \cdot \cos\left(B_1 \cdot (x + C_1)\right) + D_1$

$g(x) = A_2 \cdot \cos\left(B_2 \cdot (x + C_2)\right) + D_2$

Is it possible at all to solve this analytically? I can start doing this but I get stuck half way.

$A_1 \cdot \cos\left(B_1 \cdot (x + C_1)\right) + D_1 = A_2 \cdot \cos\left(B_2 \cdot (x + C_2)\right) + D_2$

$\Longleftrightarrow A_1 \cdot \cos\left(B_1 \cdot (x + C_1)\right) - A_2 \cdot \cos\left(B_2 \cdot (x + C_2)\right) = D_2 - D_1$

I'm not sure how to use arccosine on this expression. Therefore I'm asking for help to solve this.

Thanks in advance!

  • $\begingroup$ solve for all $x\in\mathbb{R}$ for general $A_i,B_i,C_i,D_i$? looks impossible. also, does not always exists for certain values, unless you want to solve it over complex numbers $\endgroup$
    – Lost1
    Jan 14, 2014 at 13:09
  • 1
    $\begingroup$ @Robin: I think you might have to resort to numerical approaches for a set of constants. $\endgroup$
    – Amzoti
    Jan 14, 2014 at 13:12

1 Answer 1


Substituting $\xi:=B_1(x+C_1)$ brings it to the fundamental form

$$ \cos (\xi) = p\cos (a\xi+b)+q$$

with $p=\frac{A_2}{A_1}$, $q=\frac{D_2-D_1}{A_1}$, $a=\frac{A_2}{B_1}$ and $b=\frac{C_2-C_1}{B_1}$. As far as I know, this form cannot be solved analytically in general.


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