Question

Sabendo que 3cosx + senx = -1. Verifique o valor de senx e cosx. ​

Answer 1

$\Large \sf 3cos x +sen x = -1 \\\\ 3cos x = -(1+senx) \\\\ \underline{elevando \ ao \ quadrado \ dos \ dois \ lados}: \\\\ (3cosx)^2=(-1)^2(1+senx)^2 \\\\ 9cos^2x=1+2sen x+sen^2x \\\\ \underline{Usando \ a\ rela{\c c}{\~a}o \ fundamental \ da\ trigonometria }: \\\\ \Large\boxed{\begin{array}{I}\sf sen^2x+cos^2x = 1 \to \sf cos^2x = 1-sen^2x \end{array}} \\\\\\ {Da{\'i}}:$

$\displaystyle \Large \sf 9(1-sen^2x)=1+2sen x+sen^2x \\\\ 9-9sen^2x=sen^2x+2sen x+1 \\\\ 10. sen^2x +2.sen x -8 = 0 \\\\ sen x = \frac{-2\pm\sqrt{2^2-4.10.(-8)}}{2.10 } \\\\\\ sen x= \frac{-2\pm\sqrt{4+320}}{20}\to sen x= \frac{-2\pm\sqrt{324}}{20} \\\\\\ sen x= \frac{-2\pm18}{20} \\\\\\\ \boxed{\sf sen x=-1 \ \ ; \ \ sen x = \frac{4}{5} \ }\checkmark$

Achando os possíveis valores do cosseno :

$\displaystyle \underline{\sf se\ sen x = - 1} : \\\\ \sf cos^2 x=1-sen^2x\\\\ cos^2x = 1-(-1)^2 \\\\ cos^2x = 1-1 \\\\ cos x = 0$

$\displaystyle \underline{\text{se sen x }=\frac{4}{5}}: \\\\\\ \sf cos^2x =1-\left(\frac{4}{5}\right)^2 \\\\\\ cos^2x=1-\frac{16}{25} \\\\\\ cos^2x=\frac{25-16}{25} \\\\\\ cosx =\pm\sqrt{\frac{9}{25}} \\\\\\ cos x= \frac{\pm3}{5}$

testando os valores, vemos que o positivo não convém :

$\displaystyle \sf cosx = \frac{+3}{5} \ \to\ 3\cdot \frac{3}{5}+sen x = -1 \to senx = \frac{-14}{5} \ \text{(nao conv{\'e}m)}$

Portanto :

$\Large \boxed{\begin{array}{I}\displaystyle \sf sen x = -1 \ \ ; \ \ sen x= \frac{4}{5} \\\\ \sf \ \ cos x= 0 \ \ \ \ \ ;\ \ \displaystyle cosx =\frac{-3}{5} \end{array}}\checkmark$