How many real solutions does sin 5 x − cos 5 x = c o s x 1 − s i n x 1 have? Assume sin x  = cos x (and the reciprocals must exist).

Question

How many real solutions does sin 5 x − cos 5 x = c o s x 1 ​ − s i n x 1 ​ have? Assume sin x  = cos x (and the reciprocals must exist).

TestHub · Accepted Answer

The right side is (sin x - cos x)/(sin x cos x). Cancelling the nonzero sin x - cos x, we need sin x cos x(sin⁴x + sin³x cos x + sin²x cos²x + sin x cos³x + cos⁴x) = 1. Writing t = sin x cos x ∈ [-1/2, 1/2], the equation becomes t(1 + t - t²) = 1, since sin⁴x + cos⁴x = 1 - 2t² and sin³x cos x + sin x cos³x = t. But t + t² - t³ ≤ 5/8 < 1 on this interval. Thus no solution.

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