Finding Solutions to ln(x+1) = sin^2(x) on the Interval (0,3)

Thread starter tachu101
Start date Jan 4, 2009

In summary, the conversation discusses the solution to the equation ln(X+1)=Sin^2(x) for x in the interval (0,3). It is solved by using algebraic manipulation and trigonometric identities, and the use of a graphing calculator is also mentioned. It is noted that there are two solutions in the given interval, and changing the interval can affect the solutions.

Jan 4, 2009

tachu101

Homework Statement

ln(X+1)=Sin^2(x) solve

The Attempt at a Solution

I can get three solutions 0, .964, 1.684 using the calculator. How do you solve this wo/ a calculator.

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Jan 4, 2009

jgens

Gold Member

My only thought is Newton's method.

1. What is the solution to ln(X+1)=Sin^2(x) for x in the interval (0,3)?

The solution to this equation is x = 1.078 or approximately 1.08.

2. How do you solve ln(X+1)=Sin^2(x) for x in the interval (0,3)?

To solve this equation, you can use algebraic manipulation and trigonometric identities to rewrite the equation as x = arcsin(sqrt(ln(X+1))). Then, use a calculator to find the inverse sine function of the square root of ln(X+1).

3. Can you use a graphing calculator to solve ln(X+1)=Sin^2(x) for x in the interval (0,3)?

Yes, you can use a graphing calculator to solve this equation. Graph both ln(X+1) and Sin^2(x) and find the point of intersection, which will give you the solution for x.

4. Is there more than one solution to ln(X+1)=Sin^2(x) for x in the interval (0,3)?

Yes, there are two solutions to this equation in the given interval.

5. How does changing the interval affect the solutions to ln(X+1)=Sin^2(x)?

Changing the interval can change the number and values of the solutions. In this case, the interval (0,3) was chosen because it contains the two solutions, but using a different interval may result in no solutions or different solutions.