Why is a Non-Zero Root Used in Newton's Method for Sin(x) = x^2?

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Miike012
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Question:
Use Newtons method to approximate the indicated root of the equation correct to six decimal places.

The positive root of sin(x) = x^2

The answer is ...0.876726

Why did they pick this when the obvious root is 0?

Sin(0) = (0)^2 = 0
 
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Yes, they wouldn't ask you to use Newton's Method if they only wanted the trivial root. Try to start with a first approximation that is near the root. I'd try 1.