Едно приятно тригонометрично уравнение...

[S.B.]

$$\frac{sin3x}{sin2x} = 2 .\frac{sin2x}{sin3x}$$

[Петър Евгениев]

$$\frac{sin3x}{sin2x} = 2 .\frac{sin2x}{sin3x}$$

[tex]\psi=\frac{2}{\psi}, \psi \ne 0 \Rightarrow \psi^{2}=2 \Rightarrow \psi=\pm \sqrt{2}; \frac{sin3x}{sin2x}=\sqrt{2} \cup \frac{sin3x}{sin2x}=-\sqrt{2}[/tex]
Двете имат по 4 решения, общо 8. Утре по-подробно. Лека от мен.

[Петър Евгениев]

$$\frac{sin3x}{sin2x}=\sqrt{2} \Rightarrow sin2xcosx+sinxcos2x=\sqrt{2}sin2x, x\ne k\pi \Rightarrow sinxcos2x=sin2x(\sqrt{2}-cosx)$$
$$\cancel{sinx}cos2x=2\cancel{sinx}cosx(\sqrt{2}-cosx), x\ne k\pi \Rightarrow 2cos^{2}x-1=2\sqrt{2}cosx-2cos^{2}x, cosx=\varphi \in [-1;1] $$
$$4\varphi^{2}-2\sqrt{2}\varphi-1=0 \Rightarrow \varphi_{1;2}=\frac{\sqrt{2}\pm \sqrt{6}}{4}, \varphi_{1}\notin D , \varphi_{2}=\frac{\sqrt{2}-\sqrt{4}}{4}\in D$$
$$cosx=\frac{\sqrt{2}-\sqrt{6}}{4} \Rightarrow \begin{cases} x=2k\pi-arccos\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right) \\ x=2k\pi+arccos\left(\frac{\sqrt{2}-\sqrt{6}}{4} \right) \end{cases} $$
$$\frac{sin3x}{sin2x}=-\sqrt{2} (...)long story short \rightarrow 4cos^{2}x+2\sqrt{2}cosx-1=0, x\ne k\pi \Rightarrow 4\varphi^{2}+2\sqrt{2}\varphi-1=0 \Rightarrow \varphi_{1;2}=\frac{-\sqrt{2}\pm\sqrt{6}}{4}, \varphi_{1}\notin D, \varphi_{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$$
$$cosx=\frac{\sqrt{6}-\sqrt{2}}{4} \Rightarrow \begin{cases} x=2k \pi -arccos\left(\frac{\sqrt{6}-\sqrt{2}}{4} \right) \\ x=2k \pi+arccos\left(\frac{\sqrt{6}-\sqrt{2}}{4} \right) \end{cases}$$

Скрит текст: покажи

[tex]105^\circ, 75^\circ[/tex]