This note covers solving equations that contain sine, cosine and tangent, in a given interval, using exact values, your GDC, the graph, and identities that turn an equation into a simpler one. These methods find when a repeating quantity, such as a tide height or a signal, takes a given value.
Key ideas
- A trigonometric equation usually has more than one solution in an interval, because the functions repeat. Always read the interval carefully and list every solution inside it.
- Step 1: isolate the trigonometric function, for example \(\sin x = k\).
- Step 2: find the principal value with the inverse function (\(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\)). Use exact values when you know them, or your GDC otherwise.
- Step 3: use symmetry to find the other solutions in one period, then add or subtract whole periods to cover the interval.
- For \(\sin\theta = k\): solutions are \(\theta\) and \(\pi - \theta\) (or \(180^\circ - \theta\)), plus multiples of \(2\pi\).
- For \(\cos\theta = k\): solutions are \(\theta\) and \(2\pi - \theta\) (or \(-\theta\)), plus multiples of \(2\pi\).
- For \(\tan\theta = k\): solutions are \(\theta\) plus multiples of \(\pi\).
- If \(|k| \gt 1\) for sine or cosine, there are no solutions.
Equations with a multiple or shifted angle. For an equation such as \(\cos(2x) = 0.3\), let \(u = 2x\) and find \(u\) over the widened interval. If \(0 \leq x \leq 2\pi\), then \(0 \leq u \leq 4\pi\), so you need two full periods of \(u\). Solve for \(u\), then divide by 2. For a shift such as \(\sin\left(x + \frac{\pi}{3}\right) = k\), shift the interval for \(u = x + \frac{\pi}{3}\), solve, then subtract \(\frac{\pi}{3}\).
Equations that need an identity. If the equation has two different trigonometric forms, use an identity to get one form:
- Replace \(\tan x\) by \(\frac{\sin x}{\cos x}\), or replace \(\cos^2 x\) by \(1 - \sin^2 x\), so that only one function remains.
- Replace a double angle, \(\sin 2x\) or \(\cos 2x\), so that only \(x\) remains.
- If the result is a quadratic in \(\sin x\) or \(\cos x\), factorise or use the quadratic formula (or GDC), then solve each simple equation.
- Do not divide by a trigonometric function that could be zero. Factorise instead, or you lose solutions.
Graphical method. You can also plot both sides on your GDC and find the intersections within the interval. This is useful when an equation cannot be solved algebraically, for example \(\sin x = 0.2x\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\sin\theta = \sin(\pi - \theta)\) | Second solution of a sine equation. |
| \(\cos\theta = \cos(2\pi - \theta)\) | Second solution of a cosine equation. |
| \(\tan\theta = \tan(\theta + \pi)\) | Tangent repeats every \(\pi\) (\(180^\circ\)). |
| \(\sin^2 x + \cos^2 x = 1\) | Converts between \(\sin^2\) and \(\cos^2\). |
| \(\sin 2x = 2\sin x\cos x\) | Removes a double angle from sine. |
| \(\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1\) | Removes a double angle from cosine. |
| Interval for \(u = bx + c\) | Transform the interval for \(x\) the same way before solving for \(u\). |
Worked examples
Example 1 Solve \(2\sin x = \sqrt{3}\) for \(0 \leq x \leq 2\pi\), giving exact answers.
Isolate: \(\sin x = \dfrac{\sqrt{3}}{2}\).
The principal value is \(x = \dfrac{\pi}{3}\), since \(\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}\).
The second solution in the interval is \(\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}\). Adding \(2\pi\) goes outside the interval.
Answer: \(x = \frac{\pi}{3}\) or \(x = \frac{2\pi}{3}\).
Example 2 Solve \(\cos(2x) = 0.3\) for \(0 \leq x \leq 2\pi\).
Let \(u = 2x\). Then \(0 \leq u \leq 4\pi\), which is two full periods.
Use the inverse cosine on your GDC in radian mode: \(u = \cos^{-1}(0.3) \approx 1.2661\).
The other solution in the first period is \(2\pi - 1.2661 \approx 5.0171\).
Add \(2\pi\) to each to get the solutions in the second period: \(1.2661 + 2\pi \approx 7.5493\) and \(5.0171 + 2\pi \approx 11.3003\).
Divide each by 2: \(x \approx 0.633,\ 2.51,\ 3.77,\ 5.65\).
Answer: \(x \approx 0.633,\ 2.51,\ 3.77\) or \(5.65\).
Example 3 Solve \(\cos 2x = \sin x\) for \(0 \leq x \leq 2\pi\), giving exact answers.
The equation has \(2x\) and \(x\), so replace \(\cos 2x\) with the form that uses only sine: \(1 - 2\sin^2 x = \sin x\).
Rearrange to a quadratic in \(\sin x\): \(2\sin^2 x + \sin x - 1 = 0\).
Factorise: \((2\sin x - 1)(\sin x + 1) = 0\), so \(\sin x = \dfrac{1}{2}\) or \(\sin x = -1\).
\(\sin x = \frac{1}{2}\): \(x = \frac{\pi}{6}\) or \(\pi - \frac{\pi}{6} = \frac{5\pi}{6}\).
\(\sin x = -1\): \(x = \frac{3\pi}{2}\).
Answer: \(x = \frac{\pi}{6},\ \frac{5\pi}{6}\) or \(\frac{3\pi}{2}\).
Common errors
- Giving only the principal value from the calculator and missing the other solutions in the interval.
- Solving \(\cos(2x) = k\) over the interval for \(x\) instead of the doubled interval for \(2x\), which loses half the solutions.
- Dividing both sides by \(\sin x\) or \(\cos x\), which removes the solutions where that function is zero.
- Leaving the calculator in degree mode when the interval is in radians (or the other way round).
- Writing a solution outside the interval, or including an endpoint that fails to satisfy the equation. Check each answer against the interval.
Exam tip: Sketch the graph of the function over the interval and draw the horizontal line for the value. Counting the intersections tells you how many solutions to expect.
Check yourself
- Solve \(\tan x = -2\) for \(0^\circ \leq x \leq 360^\circ\).
Answer
\(\tan^{-1}(-2) \approx -63.4^\circ\). Add \(180^\circ\): \(x \approx 116.6^\circ\). Add \(360^\circ\): \(x \approx 296.6^\circ\). - Solve \(2\cos^2 x - 3\cos x + 1 = 0\) for \(0 \leq x \leq 2\pi\).
Answer
Factorise: \((2\cos x - 1)(\cos x - 1) = 0\), so \(\cos x = \frac{1}{2}\) or \(\cos x = 1\). \(\cos x = \frac{1}{2}\) gives \(x = \frac{\pi}{3}, \frac{5\pi}{3}\). \(\cos x = 1\) gives \(x = 0, 2\pi\). So \(x = 0, \frac{\pi}{3}, \frac{5\pi}{3}, 2\pi\). - Solve \(\cos\left(x + \dfrac{\pi}{3}\right) = 0\) for \(0 \leq x \leq 2\pi\).
Answer
Let \(u = x + \frac{\pi}{3}\), so \(\frac{\pi}{3} \leq u \leq \frac{7\pi}{3}\). \(\cos u = 0\) at \(u = \frac{\pi}{2}, \frac{3\pi}{2}\) (the next one, \(\frac{5\pi}{2}\), is outside the interval). Subtract \(\frac{\pi}{3}\): \(x = \frac{\pi}{6}, \frac{7\pi}{6}\). - Solve \(\sin 2x = \sin x\) for \(0 \leq x \leq 2\pi\).
Answer
\(2\sin x\cos x - \sin x = 0\), so \(\sin x(2\cos x - 1) = 0\). \(\sin x = 0\) gives \(x = 0, \pi, 2\pi\). \(\cos x = \frac{1}{2}\) gives \(x = \frac{\pi}{3}, \frac{5\pi}{3}\). So \(x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi\).