$$\text{WTS }\exists^\infty x\in\mathbb{R},s.t. 2\sin\left(3x\right)+6\cos\left(5x\right)=0$$
Rough work $$\forall n\in\mathbb{N},\sin(n\pi)=0\wedge\cos(n\pi)=(-1)^n$$
$$\Rightarrow \forall n\pi\in\mathbb{R},2\sin(3n\pi)=0$$
$$\wedge6\cos(5n\pi)=6(-1)^{5n}$$
Proof.
Let $f:\mathbb{R\to\mathbb{R}},s.t.f(x)=2\sin\left(3x\right)+6\cos\left(5x\right)$
And $f$ is continuous
Let $a_n=n\pi,S:=\{a_n\}_{n=1}^\infty$, then
$$\forall i\in\mathbb{N},x_i\in S$$
$$f(x_i)=6\wedge f(x_{i+1})=-6$$
$$\vee f(x_i)=-6\wedge f(x_{i+1})=6$$
$$\overset{\text{IVT}}\implies\forall i\in\mathbb{N,}\exists c_i\in\mathbb{R},s.t.\min(f(x_i),f(x_{i+1}))<c_i<\max(f(x_i),f(x_{i+1}))$$
$$\Rightarrow\forall k\in\mathbb{N}, c\in \{c_1,\dots,c_k\},2\sin\left(3c\right)+6\cos\left(5c\right)=0$$
$$\Rightarrow\forall k\in\mathbb{N},\exists^{\ge k}c,2\sin\left(3c\right)+6\cos\left(5c\right)=0$$
$$\Rightarrow \exists^\infty c\in\mathbb{R},s.t. 2\sin\left(3c\right)+6\cos\left(5c\right)=0\tag*{$\square$}$$
$\dots$ Is my proof correct ? Any suggestions would be appreciated.
Also please tell me if there is a better method to prove it.
Thanks for your help.
Update:
here is some defs. for $\exists^{\ge n}$ and $\exists^\infty$
Def.
$\text{Let n $\in\mathbb{N}$, then }\exists^{\ge n}x,p(x) \text{ if and only if :(exists at least n)}$
$$\exists x_1\dots x_n \text{ s.t.}\underbrace{(p(x_1)\wedge\dots\wedge p(x_n))}_{\text{$x_1\dots x_n$ satisfy $p$}}
\wedge\underbrace{(x_1\neq x_2\dots x_1\neq x_n)\wedge\dots\wedge(x_{n-1}\neq x_n)}_{\text{$x_1\dots x_n$ are distinct}}$$
$$\Leftrightarrow\underset{i=1}{\overset{n}{\exists}} x_i,(\bigwedge_{i=1}^np(x_i))\wedge(\bigwedge_{i=1}^{n-1}(\bigwedge_{j=i+1}^nx_i\neq x_j))$$
$\exists^{\infty}x,p(x)$ if and only if:(exists infinitely many) $$\forall n\in\mathbb{N},\exists^{\ge n}x,p(x)$$
$$\Leftrightarrow \forall n\in\mathbb{N},\underset{i=1}{\overset{n}{\exists}} x_i,(\bigwedge_{i=1}^np(x_i))\wedge(\bigwedge_{i=1}^{n-1}(\bigwedge_{j=i+1}^nx_i\neq x_j))$$
Best Answer
$2\sin (3x) + 6\cos(5x)$ is the sum of two periodic functions of period $2\pi,$ making it a function with period $2\pi$
You only need to show that one zero exists.
$|2\sin 3x| \le 2$
As there are values of $x$ such that $6\cos 5x > 2$ and values where $6\cos 5x < 2$ and sine and cosine are continuous functions, there must be a zero in between (by the IVT).