Trigonometria: Relação Trigonométrica Fundamental
Enviado: 24 Set 2008, 10:23
Ssabendo que [tex3]3\text{sen}^2x+2\cos^2x=\frac{11}{4},[/tex3] com [tex3]x[/tex3] no primeiro quadrante, determine [tex3]\text{sen} x.[/tex3]
- [tex3]\text{sen}^2x+\cos^2x=1\Longrightarrow \cos^2x=1-\text{sen}^2x[/tex3]
- [tex3]3\text{sen}^2x+2\cos^2x=\frac{11}{4}\\
3\text{sen}^2x+2(1-\text{sen}^2x)=\frac{11}{4}\\
3\text{sen}^2x+2-2\text{sen}^2x=\frac{11}{4}\\
\text{sen}^2x=\frac{11}{4}-2\\
\text{sen}^2x=\frac{3}{4}\\
\text{sen}x=\pm\frac{\sqrt3}{2}[/tex3]