sin ^ 2 ( x ) + cos ^ 2 ( x ) = 1
=>
cos ^ 2 ( x ) = 1 - sin ^ 2 ( x )
y = 2 sin ( x ) - [ 1 - sin ^ 2 ( x ) ]
y = 2 sin ( x ) - 1 + sin ^ 2 x
y = sin ^ 2 x + 2 sin ( x ) - 1
Substitution sin ( x ) = t
y = t ^ 2 + 2 t - 1
x - intercept is a point on the graph where y is zero
So you must solve equation :
t ^ 2 + 2 t - 1 = 0
The solutions are :
t = sqroot ( 2 ) - 1
sin ( x ) = sqroot ( 2 ) - 1
and
t = - 1 - sqroot ( 2 ) = - [ 1 +sqroot ( 2 ) ] = - 2.41421
sin ( x ) = - 2.41421
sine cant be less of - 1 so the sotions are :
sin ( x ) = sqroot ( 2 ) - 1
x = arc sin [ sqroot ( 2 ) - 1 ]
y = 0
Coordinate of x - intercept :
arc sin [ sqroot ( 2 ) - 1 ] , 0
Y=2sinx-cos^2(x) on [0, 2pi]
How do you solve for the x-intercept
4 answers
arc sin ( x ) is inverse sine function of x
arc sin ( x ) = sin ^ - 1 ( x )
arc sin ( x ) = sin ^ - 1 ( x )
I don't understand where you got t=sqroot(2)-1 and what would the actual x-intercepts be? I'm really confused.
The solutions of quadratic equation
t ^ 2 + 2 t - 1 = 0
are :
t = sqroot ( 2 ) - 1
and
t = - 1 - sqroot ( 2 ) = - [ 1 +sqroot ( 2 ) ] = - 2.41421
Substitution
sin ( x ) = t
OR
t = sin ( x )
give :
sin ( x ) = sqroot ( 2 ) - 1
and
sin ( x ) = - 1 - sqroot ( 2 ) = - [ 1 +sqroot ( 2 ) ] = - 2.41421
Coordinate of x - intercep are :
arc sin [ sqroot ( 2 ) - 1 ] , 0
OR
0.427079 , 0
Measured in radians.
t ^ 2 + 2 t - 1 = 0
are :
t = sqroot ( 2 ) - 1
and
t = - 1 - sqroot ( 2 ) = - [ 1 +sqroot ( 2 ) ] = - 2.41421
Substitution
sin ( x ) = t
OR
t = sin ( x )
give :
sin ( x ) = sqroot ( 2 ) - 1
and
sin ( x ) = - 1 - sqroot ( 2 ) = - [ 1 +sqroot ( 2 ) ] = - 2.41421
Coordinate of x - intercep are :
arc sin [ sqroot ( 2 ) - 1 ] , 0
OR
0.427079 , 0
Measured in radians.