Question

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Answer 1

7) Pela relação fundamental trigonométrica
$sen^2x+cos^2x=1\\ sen^2x+0,3^2=1\\ sen^2x=1-0,09\\ sen^2x=0,91\\ senx= \frac{+}{-} \sqrt{0,91}$

8) Usando o mesmo procedimento
$sen^2x+cos^2x=1\\\\ senx = \frac{\sqrt{m+1}}{2} \ \ \ cosx = \frac{m}{4}\\\\ (\frac{\sqrt{m+1}}{2})^2+(\frac{m}{4})^2=1\\\\ \frac{m+1}{4}+\frac{m^2}{16}=1\\\\ \frac{4m+4}{16}+ \frac{m^2}{16}=1\\\\ m^2+4m+4=16\\\\ m^2+4m-12=0\\\\ \delta=4^2-4*1*-12\\ \delta=16+48=64\\\\ m=\frac{-4\frac{+}{-}\sqrt{64}}{2}\\\\ m=-2\frac{+}{-}4\\\\ S= \{m \in R| \ m= -6, \ m = 2 \}$