Equação Trigonométrica

por matheusfern57 Sex 13 Jan 2017, 12:26

(CEFET) - Sabendo que x satisfaz a equação :
2sen^2(x/2)+ sen(x/2) -1 = 0 no intervalo [0;pi/3] , o valor de :
cos x +tg^2x + cossec^2x é:

Gabarito: 29/6

Quem puder demonstrar a resolução agradeço!!

por **Elcioschin** Sex 13 Jan 2017, 13:26

2.sen²(x/2) + sen(x/2) - 1 = 0

Isto é uma simples equação 2º grau na variável sen(x/2)

1) Calcule as raízes sen(x/2) e veja qual delas atende

2) Calcule cos(x/2)

3) Calule senx = 2.sen(x/2).cos(x/2) e cosx

4) Calcule tg²x e cossec²x

Complete

por superaks Sex 13 Jan 2017, 13:55

Olá Matheus.

$\mathsf{2\cdot sen^2(\frac{x}{2})+\underbrace{\mathsf{sen(\frac{x}{2})}}-1=0}\\~~~~~~~~~~~~\qquad\qquad~~\mathsf{y}\\\\\mathsf{2y^2+y-1=0}\\\\\mathsf{\Delta=1^2-4\cdot2\cdot(-1)}\\\mathsf{\Delta=1+8}\\\mathsf{\Delta=9}\\\\\mathsf{y=\dfrac{-1\pm\sqrt{9}}{2\cdot2}}\\\\\\\mathsf{y^+=\dfrac{-1+3}{4}~~~~~~~~~~~y^-=\dfrac{-1-3}{4}}\\\\\boxed{\mathsf{y^+=\dfrac{1}{2}}}~~~~~\qquad\qquad\boxed{\mathsf{y^-=-1}}\mathsf{~\gets~Fora~do~intervalo}$

$\mathsf{sen(\frac{x}{2})=\frac{1}{2}\Rightarrow \frac{x}{2}=\frac{\pi}{6}\Rightarrow \boxed{\boxed{\mathsf{x=\frac{\pi}{3}}}}}\\\\\\\\\mathsf{cos(\frac{\pi}{3})+tg^2(\frac{\pi}{3})+cossec^2(\frac{\pi}{3})\Rightarrow \frac{1}{2}+\Big(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\Big)^2+\Big(\frac{1}{\frac{\sqrt{3}}{2}}\Big)^2}\\\\\\\mathsf{\frac{1}{2}+\frac{3}{\diagup\!\!\!\!4}\cdot \diagup\!\!\!\!4+1\cdot\frac{4}{3}\Rightarrow \frac{3+18+8}{6}\Rightarrow\boxed{\boxed{\mathsf{ \frac{29}{6}}}}}$

@Edit. Corrigido.