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Pure · Trigonometry

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Solving trigonometric equations

The full method for solving trig equations in a range — quadratics in sin/cos, using sin²+cos²≡1 to convert a mixed equation, and transformed arguments like sin 2x and cos(x − 30°) where the interval must change.

Pure · Trigonometry

Solving trigonometric equations

The full method for solving trig equations in a range — quadratics in sin/cos, using sin²+cos²≡1 to convert a mixed equation, and transformed arguments like sin 2x and cos(x − 30°) where the interval must change.

Why it works

Every trig equation has the same skeleton: get it into the form sin⁡(something)=k\sin(\text{something}) = k (or cos⁡\cos, or tan⁡\tan), find the first value, then use the graph's symmetry to collect all solutions in the range. Three things make them harder — each with a standard fix.

1. Quadratics in a trig function. An equation like 2sin⁡2x−sin⁡x−1=02\sin^2x - \sin x - 1 = 0 is just a quadratic in s=sin⁡xs = \sin x. Substitute, factorise (2s+1)(s−1)=0(2s + 1)(s - 1) = 0, solve for ss, then solve sin⁡x=\sin x = each value. A warning that catches people out: discard any value with ∣s∣>1|s| > 1 — sin⁡\sin and cos⁡\cos can never exceed 11, so that branch has no solutions.

2. A mix of sin and cos² (or cos and sin²). Use sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1 to turn everything into one function, which makes it a quadratic. For example 2cos⁡2x=1+sin⁡x2\cos^2x = 1 + \sin x becomes 2(1−sin⁡2x)=1+sin⁡x2(1 - \sin^2x) = 1 + \sin x, i.e. 2sin⁡2x+sin⁡x−1=02\sin^2x + \sin x - 1 = 0 — now solvable as above.

3. A transformed argument. For sin⁡2x=0.5\sin 2x = 0.5 on 0≤x≤360°0 \le x \le 360°, let u=2xu = 2x. As xx runs over [0°,360°][0°, 360°], uu runs over [0°,720°][0°, 720°] — a doubled range — so solve sin⁡u=0.5\sin u = 0.5 over 00–720°720° (which has four solutions), then halve each. For cos⁡(x−30°)\cos(x - 30°), let u=x−30°u = x - 30° and shift the range to [−30°,330°][-30°, 330°], solve, then add 30°30° back. The classic mistake is leaving the range alone, so half the solutions go missing.50100150200250300350-1-0.50.51x°yy=sin⁡2xy = \sin 2x fits two full waves into 00–360°360° — that's why an equation like sin⁡2x=0.5\sin 2x = 0.5 has twice as many solutions as sin⁡x=0.5\sin x = 0.5. Always sanity-check the count: range width ÷ period, times the number of values.