Stationary Points of x-2xsinx: 0-3pi

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The stationary points of the function f(x) = x - 2x sin(x) within the interval [0, 3π] are determined by setting the derivative, f'(x) = 1 - 2x cos(x) - 2 sin(x), equal to zero. The differentiation is correct, but solving the resulting equation analytically is not feasible. Instead, the Newton-Raphson method is recommended for numerically approximating the critical values of x.

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markosheehan
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what are the stationary points of x-2xsinx in the interval of [0,3pi]
i differentiated it and let it equal to 0 . i get when i let 1-2x cosx-2sinx equal 0 i can't solve it
 
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You've differentiated correctly, and it appears to me you will have to use a numeric root finding method, such as the Newton-Raphson method to approximate the critical values. :D
 

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