Resolver $\tan (\theta) + \tan (2\theta) = \tan (3\theta)$

Question

Encuentra la solución general de: $$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$

Mi intento: $$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$ $$\dfrac {\sin (\theta)}{\cos (\theta)}+ \dfrac {\sin (2\theta)}{\cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$ $$\dfrac {\sin (\theta+2\theta)}{\cos (\theta) \cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$

Answer 1

\begin {align} \tan ( \theta ) + \tan (2 \theta ) &= \tan (3 \theta ) \\ \tan ( \theta ) + \tan (2 \theta ) &= \tan ( \theta + 2 \theta ) \\ \dfrac { \tan ( \theta ) + \tan (2 \theta )}{1} &= \dfrac { \tan ( \theta ) + \tan (2 \theta )} {1- \tan ( \theta ) \tan (2 \theta )} \\ \end {align}

Así que, o bien $\tan (\theta) + \tan (2\theta)=0$ o $\tan (\theta) = 0$ o $\tan (2\theta)=0$

\begin {align} \tan ( \theta ) + \tan (2 \theta ) &= 0 \\ \tan ( \theta ) + \dfrac {2 \tan ( \theta )}{1 - \tan ^2( \theta )} &= 0 \\ 3 \tan ( \theta ) - \tan ^3( \theta ) &= 0 \\ \tan ( \theta ) & \in \{0, \pm \sqrt 3\} \end {align}

Así que $\theta \in \left\{ n\pi, \pm\frac 13\pi + n\pi, \pm\frac 14\pi + n\pi : n \in \mathbb Z \right\}$

Answer 2

Compruebe si uno de $\cos\theta,\cos2\theta,\cos3\theta=0$

Si no, tenemos $$\sin3\theta(\cos3\theta-\cos\theta\cos2\theta)=0$$

¿Y si $\sin3\theta=0?$

Si no, utilice $$2\cos\theta\cos2\theta=\cos3\theta+\cos\theta$$

Answer 3

Te daré una pista para que empieces, y es que $\text{tan}(a + b) = \dfrac{\text{tan}(a) + \text{tan}(b)}{1-\text{tan}(a)\text{tan}(b)}$ .