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Number of values of `p` for which equation `sin^3x+1+p^3-3psinx=0(p >0) has a root is ________

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To solve the equation \( \sin^3 x + 1 + p^3 - 3p \sin x = 0 \) for the number of values of \( p \) such that the equation has a root, we can follow these steps: ### Step 1: Rearranging the Equation We start with the given equation: \[ \sin^3 x + 1 + p^3 - 3p \sin x = 0 \] Rearranging gives us: \[ \sin^3 x + p^3 = 3p \sin x - 1 \] ### Step 2: Substituting \( y = \sin x \) Let \( y = \sin x \). The equation becomes: \[ y^3 + p^3 = 3py - 1 \] This can be rewritten as: \[ y^3 - 3py + (p^3 + 1) = 0 \] ### Step 3: Analyzing the Cubic Equation We need to analyze the cubic equation: \[ y^3 - 3py + (p^3 + 1) = 0 \] This is a cubic equation in \( y \). For this cubic equation to have at least one real root, we can use the discriminant or analyze the behavior of the function. ### Step 4: Finding the Critical Points To find the critical points, we differentiate the function: \[ f(y) = y^3 - 3py + (p^3 + 1) \] The derivative is: \[ f'(y) = 3y^2 - 3p \] Setting the derivative to zero gives: \[ 3y^2 - 3p = 0 \implies y^2 = p \implies y = \pm \sqrt{p} \] ### Step 5: Evaluating the Function at Critical Points Now we evaluate the function \( f(y) \) at the critical points \( y = \sqrt{p} \) and \( y = -\sqrt{p} \): 1. For \( y = \sqrt{p} \): \[ f(\sqrt{p}) = (\sqrt{p})^3 - 3p(\sqrt{p}) + (p^3 + 1) = p\sqrt{p} - 3p\sqrt{p} + (p^3 + 1) = -2p\sqrt{p} + (p^3 + 1) \] 2. For \( y = -\sqrt{p} \): \[ f(-\sqrt{p}) = (-\sqrt{p})^3 - 3p(-\sqrt{p}) + (p^3 + 1) = -p\sqrt{p} + 3p\sqrt{p} + (p^3 + 1) = 2p\sqrt{p} + (p^3 + 1) \] ### Step 6: Conditions for Roots For the cubic equation to have at least one real root, one of these evaluations must be zero or change signs. 1. Set \( f(\sqrt{p}) = 0 \): \[ -2p\sqrt{p} + (p^3 + 1) = 0 \implies p^3 + 1 = 2p\sqrt{p} \] 2. Set \( f(-\sqrt{p}) = 0 \): \[ 2p\sqrt{p} + (p^3 + 1) = 0 \implies p^3 + 1 = -2p\sqrt{p} \] ### Step 7: Finding the Values of \( p \) We need to find the values of \( p \) such that these conditions hold true. After solving these equations, we find that the only value of \( p \) that satisfies the conditions and is greater than zero is \( p = 1 \). ### Conclusion Thus, the number of values of \( p \) for which the equation has a root is: \[ \boxed{1} \]

To solve the equation \( \sin^3 x + 1 + p^3 - 3p \sin x = 0 \) for the number of values of \( p \) such that the equation has a root, we can follow these steps: ### Step 1: Rearranging the Equation We start with the given equation: \[ \sin^3 x + 1 + p^3 - 3p \sin x = 0 \] Rearranging gives us: ...
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