Question

Y=2sinx-cos^2(x) on [0, 2pi]
How do you solve for the x-intercept

Answer 1

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

Answer 2

arc sin ( x ) is inverse sine function of x

arc sin ( x ) = sin ^ - 1 ( x )

Answer 3

I don't understand where you got t=sqroot(2)-1 and what would the actual x-intercepts be? I'm really confused.

Answer 4

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.