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Home » Thus, the number of distinct sequences is $\boxed326592$. - AMAZONAWS

Thus, the number of distinct sequences is $\boxed326592$. - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$Thus, the number of distinct sequences is $\boxed{326592}$.$

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📌 Solution: Expand $ (\sin x + 2\cos x)^2 = \sin^2 x + 4\sin x \cos x + 4\cos^2 x $. Add $ \sin^2 x $: total $ 2\sin^2 x + 4\sin x \cos x + 4\cos^2 x $. Simplify using identities: $ 2(1 - \cos^2 x) + 2\sin 2x + 4\cos^2 x = 2 + 2\cos^2 x + 2\sin 2x $. Let $ u = \cos^2 x $, $ \sin 2x = 2\sin x \cos x $. Alternatively, rewrite original expression as $ \sin^2 x + 4\sin x \cos x + 4\cos^2 x + \sin^2 x = 2\sin^2 x + 4\sin x \cos x + 4\cos^2 x $. Let $ f(x) = 2\sin^2 x + 4\sin x \cos x + 4\cos^2 x $. Use $ \sin^2 x = rac{1 - \cos 2x}{2} $, $ \cos^2 x = rac{1 + \cos 2x}{2} $, $ \sin x \cos x = rac{\sin 2x}{2} $:

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Since \( x > 0 \), \( x = rac{ -35 + \sqrt{2041} }{4} \)
\( rac{1}{2} (2a + (n-1)d) = 3n + 5 \)

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