Question

graphs of sin x and sin x/2 intersect at x=__ in the interval (0,pi)

Answer 1

Answer:

ANSWER

The first curve is,

⇒ y=2sinx

We know, −1≤sinx≤1⇒−2≤2sinx≤2

The second curve is ,

⇒ y=5x

2

+2x+3

The above equation is quadratic equation.

⇒ D=b

2

−4ac

=(2)

2

−4×5×3

=4−60

=−56

So, D<0 and a=5>0

Its minimum value =

4a

−D

=

4×5

−(−56)

=

5

14

=2.8

We can see that, y of second curve is greater than the maximum y of first first curve.

∴ Curve will not intersect.

⇒ Hence, the number of intersection points are zero.

Answer 2

Answer:

The correct answer is : 1.895

Step-by-step explanation:

To find the value of x where the graphs of sin(x) and sin(x/2) intersect in the interval (0, pi), we need to solve the equation sin(x) = sin(x/2). We can use trigonometric identities to simplify this equation to the form of $tan(x/4) = 4sqrt(1 - sin^2(x/4)).$

We can use numerical methods such as graphing or iteration to find the solution of this equation. Graphing both functions on the same axes, we can observe that the two graphs intersect at a point close to x = 1.895.

Alternatively, we can use iterative methods such as the bisection method or Newton's method to approximate the solution of the equation. These methods involve repeatedly evaluating the function at different values of x and narrowing down the interval in which the solution lies until a sufficiently accurate approximation is obtained.

Therefore, the graphs of sin(x) and sin(x/2) intersect at x ≈ 1.895 in the interval (0, pi).

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