All the solutions
Tonight's idea
On the unit circle, \(\sin x\) is the height of the point you reach after turning through \(x\) (where you met this). So \(\sin x = 0.5\) asks: where is the circle at height \(\tfrac12\)? A horizontal line cuts a circle twice.
A whole turn brings you back to the same point, so there are infinitely many solutions:
\(x = 30° + 360°n\) or \(x = 150° + 360°n\), where \(n\) is any whole number: positive, negative or zero.
The same three steps solve every equation of this kind.
1. One answer. Get one angle \(\alpha\) from the exact values or the calculator's inverse key.
2. Its partner. For sine, \(180° - \alpha\) (same height). For cosine, \(-\alpha\) (same \(x\)-coordinate). Tangent repeats every half turn, so all its solutions are \(\alpha + 180°n\).
3. Whole turns. Add \(360°n\) to both, then keep the solutions in the interval the question asks for.
In radians a whole turn is \(2\pi\), so the callout reads \(x = \tfrac{\pi}{6} + 2\pi n\) or \(\tfrac{5\pi}{6} + 2\pi n\).
Watch one: \(\sin 2x = \tfrac{\sqrt3}{2}\) for \(0° \le x \lt 360°\)
Solve for the whole angle \(2x\) first. One answer is \(60°\). Its partner is \(120°\). Add whole turns:
\[2x = 60° + 360°n \quad\text{or}\quad 2x = 120° + 360°n.\]
Only now divide by 2: \(x = 30° + 180°n\) or \(x = 60° + 180°n\). In the interval, \(n = 0\) and \(n = 1\) give
\[x = 30°,\; 60°,\; 210°,\; 240°.\]
Four solutions, because \(2x\) goes round twice while \(x\) goes round once. Halving \(60°\) and \(120°\) before adding the turns would have lost two of them.
Drag the point round and watch its height: mathdetours.com/demos/unitcircle.html.
Your turn
- Find every solution with \(0° \le x \lt 360°\), without a calculator: (a) \(\sin x = -\tfrac{\sqrt3}{2}\), (b) \(\tan x = \sqrt3\).
- Use a calculator. Write down all the solutions of \(\cos x = 0.3\), to one decimal place in degrees. Then list the ones between \(-360°\) and \(360°\).
- Solve \(2\sin^2 x - \sin x - 1 = 0\) for \(0 \le x \lt 2\pi\), in radians. Treat it as a quadratic in \(\sin x\) first.
- To solve \(\tan x = 2\sin x\) for \(0° \le x \lt 360°\), a student writes \(\tan x\) as \(\tfrac{\sin x}{\cos x}\), divides both sides by \(\sin x\), and gets \(\cos x = \tfrac12\), so \(x = 60°\) or \(300°\). Two solutions are missing. Find them, and say which step lost them.
Stretch: a wave and a line
Work in radians. How many solutions does \(\sin x = \dfrac{x}{10}\) have? Sketch the wave and the line on the same axes and count the crossings. You do not need to find them. (Useful: \(3\pi \approx 9.42\).)
Answers
- (a) \(\sin 60° = \tfrac{\sqrt3}{2}\), and sine is negative in quadrants III and IV: \(x = 240°\) or \(300°\). (b) \(\tan 60° = \sqrt3\), and tangent repeats every \(180°\): \(x = 60°\) or \(240°\).
- \(\alpha = \arccos 0.3 \approx 72.5°\), partner \(-72.5°\). All solutions: \(x \approx \pm 72.5° + 360°n\). Between \(-360°\) and \(360°\): \(-287.5°,\; -72.5°,\; 72.5°,\; 287.5°\).
- \((2\sin x + 1)(\sin x - 1) = 0\), so \(\sin x = 1\) or \(\sin x = -\tfrac12\). The first gives \(x = \tfrac{\pi}{2}\). The second gives \(x = \tfrac{7\pi}{6}\) or \(\tfrac{11\pi}{6}\). Three solutions.
- Dividing by \(\sin x\) is only allowed when \(\sin x \neq 0\). When \(\sin x = 0\), both sides of the original equation are 0, so \(x = 0°\) and \(x = 180°\) are solutions too. Safer: move everything to one side and factorise, \(\sin x\left(\tfrac{1}{\cos x} - 2\right) = 0\). All four: \(0°, 60°, 180°, 300°\).
Stretch. Seven. The line leaves the range of sine when \(|x| > 10\), so only \(-10 \le x \le 10\) matters. \(x = 0\) is one solution. For \(x > 0\) the sine curve has humps above the axis on \((0, \pi)\) and \((2\pi, 3\pi)\). At 0 the wave is steeper than the line, so it starts above it and crosses back once before \(\pi\). On the second hump the line sits at about 0.63 to 0.94, below the peak of 1, so it cuts the hump twice. That is 3 positive solutions. Replacing \(x\) by \(-x\) changes the sign of both sides, so each positive solution has a negative twin: \(3 + 3 + 1 = 7\).