MathLabs

Problem 3

Solve the equation cos⁡nx−sin⁡nx=1\cos^n x-\sin^n x=1, where nn is a given positive integer.
Step 3 of 6: Restrict odd nn on the upper half-period
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n odd, x∈[0,π]:sin⁡nx≥0⟹cos⁡nx−sin⁡nx≤1n\text{ odd},\ x\in[0,\pi]:\quad \sin^n x\ge0\Longrightarrow\cos^n x-\sin^n x\le1
Detailed analysis

For odd nn and x∈[0,π]x\in[0,\pi], sin⁡x≥0\sin x\ge0, so sin⁡nx≥0\sin^n x\ge0 and cos⁡nx≤1\cos^n x\le1. Equality in the equation can therefore occur only when sin⁡x=0\sin x=0 and cos⁡x=1\cos x=1, namely x=0x=0 modulo 2π2\pi.