Question

4 sen ^2 - 8 sen + 3 =0 con forma cuadratica x=-b± √b^2 - 4ac/2 a

Answer 1

4 sen ^2 - 8 sen + 3 =0 con forma cuadratica x=-b± √b^2 - 4ac/2 a

$4\ sen ^2 \alpha - 8\ sen \alpha + 3 = 0 \qquad sustituimos \to x= sen \alpha \\ \\ 4x^2 - 8x + 3 = 0 \\ \\ \\ x_{1\ y\ 2}= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\qquad\qquad a=4 \quad b= -8\quad c= 3 \\ \\ \\ x_{1\ y\ 2}= \dfrac{8\pm \sqrt{(-8)^2-4(4)(3)} }{2(4)} \\ \\ \\ x_{1\ y\ 2}= \dfrac{8\pm \sqrt{64-48} }{8} \\ \\ \\ x_{1\ y\ 2}= \dfrac{8\pm \sqrt{16} }{8} \\ \\ \\ x_{1\ }= \dfrac{8+4 }{8}\qquad\qquad \qquad\qquad x_{2\ }= \dfrac{8-4 }{8}$

$x_{1\ }= \dfrac{12 }{8}\qquad\qquad \qquad\qquad x_{2\ }= \dfrac{4 }{8} \\ \\ \\ x_{1\ }= \dfrac{3}{2}\qquad\qquad \qquad\qquad x_{2\ }= \dfrac{1 }{2} \\ \\ x= sen \alpha \\ \\ sen \alpha _1= \dfrac{3}{2}\to NO \ hay\ ningun\ valor \ de \ \alpha \\ \\ \\ sen \alpha_2= \dfrac{1}{2}\to \alpha = arc\ seno \ \dfrac{1}{2}\to \boxed { \alpha = 30\º }$

Espero que te sirva, salu2!!!!