# Find the exact solutions on the interval [0,pi].

2sin2x = sqrt(x)

kristie, this problem is somewhere between a calculus problem and an algebra one. I'm not sure what you're taking. If you see that the eq. is = to 4(sin2x)^2=x , then you might see where they're equal graphically. I think they want you to make use of the lim sinX / X for this exercise too.

9 months ago

## To solve the equation 2sin^2x = sqrt(x), we need to find the exact solutions on the interval [0, pi].

Let's break down the problem step by step:

Step 1: Rewrite the equation using trigonometric identities.
We know that sin^2x = (1/2)(1-cos2x) and sqrt(x) = x^(1/2). So, we can rewrite the given equation as:
2(1/2)(1-cos2x) = x^(1/2)
Simplifying further:
1 - cos2x = x^(1/2)

Step 2: Solve for cos2x.
Rearranging the equation:
1 - x^(1/2) = cos2x
Using the identity cos(2Î¸) = 1 - 2sin^2(Î¸), we can substitute 2x in place of Î¸:
cos2x = 1 - 2sin^2(x)

Step 3: Substitute cos2x in the equation.
1 - 2sin^2(x) = 1 - x^(1/2)
Rearranging further:
2sin^2(x) = x^(1/2)

Step 4: Set up the equation for solving.
Now, we have:
2sin^2(x) = x^(1/2)

Step 5: Solve the equation using calculus.
To find the exact solutions on the interval [0, pi], we can use calculus techniques:

Make the observation that both sides are equal to zero at x = 0.

Differentiate both sides of the equation:
d/dx (2sin^2(x)) = d/dx (x^(1/2))
4sin(x)cos(x) = (1/2)x^(-1/2)

Solve for x using the derivative:
4sin(x)cos(x) = (1/2)x^(-1/2)

This equation does not have an analytical solution, so numerical methods or technology (like a graphing calculator or software) may be used to approximate the solutions.

In summary, the equation 2sin^2x = sqrt(x) can be solved by rewriting it, and using calculus techniques to find the exact solutions on the interval [0, pi].