$\sin(x)+p\cos(x)=2p, p\in\mathbb{R}$ has solutions for $p=?$

Question

This equation has solutions for $p=?$

$$\sin(x)+p\cos(x)=2p, p\in\mathbb{R}$$

My try: I divided the equation by $\sqrt{1+p^{2}}$ and I got $x=2k\pi+-arccos(\frac{1}{\sqrt{1+p^{2}}})+arctan(\frac{1}{p})$

I put the condition $1+p^{2}\geq0$ but I my answer is wrong.The right answer is $p\in\left [ -\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right ]$

Some ideas?

Answer 1

$\sin x+p\cos x=\sqrt{p^2+1}\sin(x+\alpha)$ for some $\alpha$.

What you need is $|2p|\le\sqrt{p^2+1}$

Answer 2

Use Weierstrass substitution

$$\dfrac{2t}{1+t^2}+p\cdot\dfrac{1-t^2}{1+t^2}=2p\iff3p t^2-2t+p=0$$ where $t=\tan\dfrac x2$ which is real

So, the discriminant must be $\ge0$

i.e., $2^2\ge4\cdot3p\cdot p\iff3p^2\le1$

Can you take it from here?