Analytical solution to $A \cos (x + \alpha) - B \cos(3x + \beta) = 0$

Question

I would like to know if the equation

$$A \cos (x + \alpha) - B \cos(3x + \beta) = 0 \qquad\text{with}\qquad A > 0,\; B >0$$

can be solved analytically to find its roots on the variable $x$. $A$, $B$, $\alpha$ and $\beta$ are given parameters. All quantities are real numbers.

If the equation has no analytical solution, I would like to understand why not, and I would appreciate suggestions on simple methods on how to find the roots.

Answer 1

We start here (with $a$ and $b$ for $\alpha$ and $\beta$, to save typing effort): $$A \cos(x+a) = B \cos(3x + b)$$ Expanding, and writing $c$ and $s$ for $\cos x$ and $\sin x$, we get $$A (\; c \cos a - s \sin a \;) = B (\; c^3 \cos b - 3 c^2 s \sin b - 3 c s^2 \cos b + s^3 \sin b \;)$$ We can reduce the powers of sine by replacing $s^2 = 1 - c^2$: $$A (\; c \cos a - s \sin a \;) = B (\; 4 c^3 \cos b - 3 c \cos b + s \sin b - 4 c^2 s \sin b \;)$$ Isolating $s$: $$s(\; A \sin a + B \sin b - 4 c^2 \sin b \;) = c (\; A \cos a - 4 B c^2 \cos b + 3 B \cos b \;)$$ Squaring, again replacing $s^2 = 1 - c^2$, and tidying-up a bit: $$\begin{align} 0 &= 16 B^2 c^6 \\ &- 8 B c^4 \;(\; 3 B + A \cos(a-b) \;) \\ &+ c^2 \;(\; A^2 + 9 B^2 + 6 A B \cos a \cos b + 10 A B \sin a \sin b \;) \\ &-\;(\;A \sin a + B \sin b\;)^2 \end{align}$$

This cubic in $c^2$ "can" be solved explicitly, but Mathematica's brute-force solution is not pretty. For a taste, here's one root:

$$\frac{4 A^2 B^2 \cos ^2(a-b)+6 B^2 \sqrt[3]{8 A^3 B^3 \cos ^3(a-b)-9 A B^3 \left(A^2+3 B^2\right) \cos (a-b)-18 A^2 B^4 \cos (2 a)+3 \left(\sqrt{3} \sqrt{B^6 (A \sin (a-b)-B \sin (2 b))^2 \left(8 A^3 B \cos (3 a-b)+A^4+18 A^2 B^2-27 B^4\right)}+3 A^2 B^4 \cos (2 b)-9 B^6 \cos (2 b)\right)}+2 A B \cos (a-b) \sqrt[3]{8 A^3 B^3 \cos ^3(a-b)-9 A B^3 \left(A^2+3 B^2\right) \cos (a-b)-18 A^2 B^4 \cos (2 a)+3 \left(\sqrt{3} \sqrt{B^6 (A \sin (a-b)-B \sin (2 b))^2 \left(8 A^3 B \cos (3 a-b)+A^4+18 A^2 B^2-27 B^4\right)}+3 A^2 B^4 \cos (2 b)-9 B^6 \cos (2 b)\right)}+\left(8 A^3 B^3 \cos ^3(a-b)-9 A B^3 \left(A^2+3 B^2\right) \cos (a-b)-18 A^2 B^4 \cos (2 a)+3 \left(\sqrt{3} \sqrt{B^6 (A \sin (a-b)-B \sin (2 b))^2 \left(8 A^3 B \cos (3 a-b)+A^4+18 A^2 B^2-27 B^4\right)}+3 A^2 B^4 \cos (2 b)-9 B^6 \cos (2 b)\right)\right)^{2/3}+6 A B^3 \cos (a+b)-3 A^2 B^2+9 B^4}{12 B^2 \sqrt[3]{8 A^3 B^3 \cos ^3(a-b)-9 A B^3 \left(A^2+3 B^2\right) \cos (a-b)-18 A^2 B^4 \cos (2 a)+3 \left(\sqrt{3} \sqrt{B^6 (A \sin (a-b)-B \sin (2 b))^2 \left(8 A^3 B \cos (3 a-b)+A^4+18 A^2 B^2-27 B^4\right)}+3 A^2 B^4 \cos (2 b)-9 B^6 \cos (2 b)\right)}}$$

Some simplification is possible. (For instance, we can move some $B^6$ and $(A \sin (a-b)-B \sin (2 b))$ factors out of some square roots.) But whether this can be wrestled into a useful form ... I can't say for sure, but I'm not optimistic.

Answer 2

Whenever Mathematica recognizes an equation allowing no analytic solution (with some specific definition as to what that means in Mathematica), it just spits the output exactly the same as the input.

btw, mine is Mathematica 10.

Answer 3

REMARK.-For the particular case where the constants are such that $\beta=3\alpha$ you have solution. In fact, $$A\cos(x+\alpha)=B\cos(3(x+\alpha))=4B\cos^3(x+\alpha)-3B\cos (x+\alpha)$$ Hence $$\begin{cases}\cos(x+\alpha)=0\\\cos(x+\alpha)=\pm\sqrt{\frac{A+3B}{4B}}\end{cases}$$ Explicity, for example,$$x=\frac{(2k+1)\pi}{2}-\alpha$$

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In the general case it would be useful a parameter $t$ as follows: $$\frac{\cos(3x+\beta)}{\cos(x+\alpha)}=\frac AB\Rightarrow\begin{cases}\cos(3x+\beta)=At\\\cos(x+\alpha)=Bt\end{cases}$$ Hence each solution of $$3\arccos(Bt)-\arccos(At)=3\alpha-\beta$$ gives a solution for $x$

(According the values of $A,B,\alpha,\beta$, is there a unique solution?, none solution? )