Estude! Com Prof. Caju

Determine os valores de [tex3]\theta \in [0,\pi][/tex3] para os quais o ponto [tex3]P[/tex3] de coordenadas [tex3](2\cos2\theta, 3\text{sen}\theta)[/tex3] pertença à elipse de equação [tex3]\frac{x^2}{4}+ \frac{y^2}{3}=1.[/tex3]

Se o ponto [tex3](2\cos2\theta, 3\text{sen}\theta)[/tex3] pertence à elipse, então:

• [tex3]\begin{array}{rl} \frac{x^2}{4}+ \frac{y^2}{3}=1&\Longrightarrow \frac{(2\cos2\theta)^2}{4}+ \frac{(3\text{sen}\theta)^2}{3}=1\\
  &\Longrightarrow \cos^22\theta+3\text{sen}^2\theta=1 \\
  &\Longrightarrow\cos^22\theta+\frac{3(1-\cos2\theta)}{2}=1 \\
  &\Longrightarrow2\cos^22\theta- 3\cos2\theta+1=0 \\
  &\Longrightarrow \begin{cases}\cos2\theta =1\Longrightarrow \cos2\theta =\cos0\Longrightarrow \theta =k\pi,\,k\in\mathbb{Z}\\ \text{ ou }\\ \cos2\theta =\frac{1}{2}\Longrightarrow\cos2\theta =\cos \frac{\pi}{3}\Longrightarrow\theta =\pm \frac{\pi}{6}+k\pi,\,k\in\mathbb{Z} \end{cases}
  \end{array}[/tex3]

Logo, [tex3]\theta\in\left\{0,\frac{\pi}{6},\frac{5\pi}{6},\pi\right\}.[/tex3]


[tex3]\diamond \text{ sen}\theta=\sqrt{\frac{1-\cos2\theta}{2}}\Longrightarrow \text{sen}^2\theta=\frac{1-\cos2\theta}{2}[/tex3]