Complexity of Irrationals

Geometry Level 3

Let a a be a irrational number . Find the solution to the equation 1 + sin ⁡ 2 ( a x ) = cos ⁡ x 1+ \sin^2 (ax) = \cos x .

x = 2 n π x=2nπ x = n π a x = \frac {n\pi}a x = 0 x=0 None of these choices

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1 solution

Chew-Seong Cheong
Jul 18, 2016

1 + sin ⁡ 2 ( a x ) = cos ⁡ x 1 + 1 2 ( 1 − cos ⁡ ( 2 a x ) ) = cos ⁡ x 3 2 − 1 2 cos ⁡ ( 2 a x ) = cos ⁡ x ⟹ cos ⁡ x + 1 2 cos ⁡ ( 2 a x ) = 3 2 \begin{aligned} 1+\sin^2 (ax) & = \cos x \\ 1 + \frac 12 \left(1-\cos (2ax) \right) & = \cos x \\ \frac 32 - \frac 12 \cos (2ax) & = \cos x \\ \implies \cos x + \frac 12 \cos (2ax) & = \frac 32 \end{aligned}

The LHS = = RHS, when cos ⁡ x = cos ⁡ ( 2 a x ) = 1 \cos x = \cos (2ax) = 1 . Since a a is irrational, the only solution is when x = 0 \boxed{x=0} .

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