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Find the number of solutions in [0, 2π) of the equation sin⁴x + cos⁴x…
Question
Find the number of solutions in [0, 2π) of the equation sin⁴x + cos⁴x = sin x · cos x.
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Step-by-step solution
Use sin⁴x + cos⁴x = (sin²x + cos²x)² − 2 sin²x cos²x = 1 − (1/2) sin²(2x), and sin x cos x = (1/2) sin 2x.
Let s = sin 2x. The equation becomes 1 − s²/2 = s/2, i.e. 2 − s² = s, or s² + s − 2 = 0, factoring as (s + 2)(s − 1) = 0. Since |s| ≤ 1, only s = 1 is valid.
sin 2x = 1 ⇒ 2x = π/2 + 2kπ ⇒ x = π/4 + kπ. In [0, 2π) this gives x = π/4 and x = 5π/4 — 2 solutions.
Final answer2
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