WEBVTT

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Here is the problem: 2sin(x) squared minus sin(x) minus 1 equals 0.

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A periodic equation in x: it has infinitely many solutions, one family for each solution in a single period.

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Start with the equation.

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Rewrite as f(x) equals 0.

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The left side repeats every 2pi, so find the solutions in one period. f(x plus 2pi) equals f(x) for every x, so every solution is one of the solutions in [0, 2pi) plus a whole number of periods.

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Solutions in one period [0, 2pi).

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Add any whole number of periods.

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So the answer is x equals pi/2 plus 2pik, x equals 7pi/6 plus 2pik, x equals 11pi/6 plus 2pik (k any integer).

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Each solution in one period was found numerically, matched to an exact value and confirmed by substitution; adding whole periods gives all solutions.

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x equals 1.5707963: left side equals 0, right side equals 0, which checks out.

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x equals 3.6651914: left side equals 0, right side equals 0, which checks out.

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x equals 5.7595865: left side equals 0, right side equals 0, which checks out.