Thus $ \sin 3x \sin x = -1 $ is achievable (e.g., $ x = rac{3\pi}{2} $). But we want **maximum** of $ \sin 3x \sin x $, not minimum.

["Maximizing the Product: How to Achieve the Maximum Value of ( \sin 3x \sin x )", "Understanding the behavior of trigonometric functions is essential in both pure mathematics and applied sciences. One intriguing problem is determining the maximum value of the expression ( \sin 3x \sin x ), particularly exploring whether and when this product reaches its peak—ideally, the maximum possible value rather than just minimums.", "In this article, we analyze the equation and identity:", "[\n\sin 3x \cdot \sin x\n]", "We aim to find the maximum value of this product and confirm whether it is achievable, citing specific values of ( x ) such as ( x = \frac{3\pi}{2} ) as possible candidates. Beyond that, we clarify the current limited maximum and discuss how trigonometric properties constrain its attainability.", "---", "### The Expression: ( \sin 3x \sin x )", "Using a classical trigonometric identity, the product-to-sum formula gives:", "[\n\sin A \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right]\n]", "Let ( A = 3x ), ( B = x ), then:", "[\n\sin 3x \sin x = \frac{1}{2} \left[ \cos(2x) - \cos(4x) \right]\n]", "So,", "[\nf(x) = \sin 3x \sin x = \frac{1}{2} \left[ \cos 2x - \cos 4x \right]\n]", "Our goal is to maximize ( f(x) ).", "---", "### Finding the Maximum Value", "We want to maximize:", "[\nf(x) = \frac{1}{2} \left( \cos 2x - \cos 4x \right)\n]", "Since ( \cos \ heta \in [-1, 1] ), the maximum of ( \cos 2x - \cos 4x ) occurs when ( \cos 2x ) is large and ( \cos 4x ) is small.", "Let’s explore particular values.", "#### Test: ( x = \frac{3\pi}{2} )", "At ( x = \frac{3\pi}{2} ):", "- ( 2x = 3\pi \Rightarrow \cos 2x = \cos(3\pi) = \cos(\pi + 2\pi) = -1 )\n- ( 4x = 6\pi \Rightarrow \cos 4x = \cos(6\pi) = \cos(0) = 1 )", "Then:", "[\nf\left( \frac{3\pi}{2} \right) = \frac{1}{2} \left( -1 - (1) \right) = \frac{-2}{2} = -1\n]", "This gives the minimum, not the maximum.", "---", "### When Is ( \sin 3x \sin x ) Maximum?", "We now seek the actual maximum. Since ( \cos 2x \leq 1 ) and ( -\cos 4x \leq 1 ), the maximum of ( \cos 2x - \cos 4x ) occurs when both terms are as large as possible—ideally:", "- ( \cos 2x = 1 \Rightarrow 2x = 2k\pi \Rightarrow x = k\pi )\n- ( \cos 4x = -1 \Rightarrow 4x = \pi + 2k\pi \Rightarrow x = \frac{\pi}{4} + \frac{k\pi}{2} )", "But no single ( x ) makes both equalities true simultaneously. So we must maximize carefully.", "Let’s use calculus to find the maximum rigorously.", "---", "### Using Calculus to Maximize ( f(x) = \frac{1}{2} (\cos 2x - \cos 4x) )", "Compute derivative:", "[\nf'(x) = \frac{1}{2} \left( -2\sin 2x + 4\sin 4x \right) = -\sin 2x + 2\sin 4x\n]", "Set ( f'(x) = 0 ):", "[\n2\sin 4x = \sin 2x\n]", "Use identity: ( \sin 4x = 2\sin 2x \cos 2x ), so:", "[\n2(2\sin 2x \cos 2x) = \sin 2x \Rightarrow 4\sin 2x \cos 2x - \sin 2x = 0\n]", "Factor:", "[\n\sin 2x (4\cos 2x - 1) = 0\n]", "Solutions:", "1. ( \sin 2x = 0 \Rightarrow 2x = k\pi \Rightarrow x = \frac{k\pi}{2} )\n2. ( 4\cos 2x - 1 = 0 \Rightarrow \cos 2x = \frac{1}{4} )", "---", "### Evaluate at Critical Points", "Case 1: ( x = \frac{k\pi}{2} )", "- ( \cos 2x = \cos(k\pi) = (-1)^k )\n- ( \cos 4x = \cos(2k\pi) = 1 )", "Then:", "[\nf(x) = \frac{1}{2} \left( (-1)^k - 1 \right)\n]", "- For even ( k ), ( (-1)^k = 1 \Rightarrow f(x) = 0 )\n- For odd ( k ), ( f(x) = \frac{1}{2}(-1 - 1) = -1 )", "Maximum here: ( 0 )", "Case 2: ( \cos 2x = \frac{1}{4} )", "Then ( \sin^2 2x = 1 - \cos^2 2x = 1 - \frac{1}{16} = \frac{15}{16} \Rightarrow \sin 2x = \pm \frac{\sqrt{15}}{4} )", "Now compute ( \cos 4x ) using double angle:", "[\n\cos 4x = 2\cos^2 2x - 1 = 2\left(\frac{1}{16}\right) - 1 = \frac{2}{16} - 1 = -\frac{14}{16} = -\frac{7}{8}\n]", "So:", "[\nf(x) = \frac{1}{2} \left( \frac{1}{4} - \left(-\frac{7}{8}\right) \right) = \frac{1}{2} \left( \frac{1}{4} + \frac{7}{8} \right) = \frac{1}{2} \left( \frac{2 + 7}{8} \right) = \frac{1}{2} \cdot \frac{9}{8} = \frac{9}{16} = 0.5625\n]", "Thus, the maximum value is ( \frac{9}{16} ), achieved when ( \cos 2x = \frac{1}{4} ).", "---", "### Is ( \sin 3x \sin x = \frac{9}{16} ) Achievable? Yes.", "Let ( \cos 2x = \frac{1}{4} ). Choose ( x = \frac{1}{2} \arccos\left( \frac{1}{4} \right) ), which is a valid real solution.", "At such ( x ), the expression reaches ( \frac{9}{16} ), the global maximum.", "The value ( x = \frac{3\pi}{2} ) gives ( -1 ), which is the minimum, not the maximum.", "---", "### Conclusion", "The maximum of ( \sin 3x \sin x ) is ( \frac{9}{16} ), achieved when ( \cos 2x = \frac{1}{4} ). While values like ( x = \frac{3\pi}{2} ) lie on the function’s domain, they yield the minimum. The peak is carefully balanced—near where both sines are moderately positive, but not fully aligned due to trigonometric constraints.", "Thus, maximizing ( \sin 3x \sin x ) is not only possible but mathematically precise, reaching up to ( \frac{9}{16} ), a value confirmed by both algebraic identities and calculus.", "---", "Key Takeaways:", "- The maximum of ( \sin 3x \sin x ) is ( \frac{9}{16} ).\n- It is achievable for specific ( x ), such as ( x = \frac{1}{2} \arccos\left( \frac{1}{4} \right) ).\n- ( x = \frac{3\pi}{2} ) gives a minimum, not maximum.\n- This problem illustrates the power of product-to-sum identities and optimization via calculus in trigonometry.", "---", "Keywords:\n( \sin 3x \sin x ) maximum, maximize ( \sin 3x \sin x ), find max of ( \sin 3x \sin x ), trigonometric maximum, maximum value of ( \sin 3x \sin x ), solution to ( \sin 3x \sin x = \max )", "SEO Meta Description:\nMaximize ( \sin 3x \sin x ) using trigonometric identities and calculus. Find the true maximum value ( \frac{9}{16} ), achievable at specific ( x ), not at ( \frac{3\pi}{2} ), revealing the actual peak of this fascinating expression."]