Respuesta :

2senx -cos²x = 0    cos²x = 1-sen²x  remplazamos
2senx -( 1 - sen²x) = 0
2senx -1 +sen²x =0
sen²x +2senx -1 =0
aplicamos

sen x =[ -b +-√(b²) -4ac]/2a
a = 1  b = 2    c = -1

senx = [ -2 + - √2² -4(1)(-1)]/2
senx = [-2 + - √4+4]/2
senx = [ -2 + - √8 ]/2
senx = [-2+ - 2√2]/2
senx = 2(-1 +- √2)/2
senx = -1 +√2
senx = -1+1,4142
senx = 0,4142
x = arcsen 0,4142
x = 24,46°

senx = -1 -√2
x = arcsen -2,4142  no existe


cos²x -3 sen²x =0          como sen²x = 1- cos²x  remplazo
cos²x -3( 1-cos²x) =0
cos²x -3 +3cos²x = 0
4cos²x -3 = 0
cos²x = 3/4
cosx = +-√3/4
cosx = + - √0,75
cosx = + - 0,866
cosx = 0,866
x = arccos0,866
x = 30°

cosx = - 0,866
x = arccos-0,866
x = 150°


pendiente(m)
p1( -5 , -4)    p2 (2,5)

ecuacion
m = (y2 -y1)/(x2 - x1)
m =[ 5 -(-4)]/[(2 -(-5)
m = (5+4)/(2+5)
m = 9/7

ecuacion