One answer from the calculator, the rest from symmetry.
Your calculator gives one solution, the principal value. The graph's symmetry gives every other solution in the interval, even when the angle is \(2x\) or \(x - 30^\circ\), or the equation needs an identity first.
Find the principal value \(\alpha\) with \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\). Then \(\sin x = k\) also has \(180^\circ - \alpha\); \(\cos x = k\) also has \(360^\circ - \alpha\); \(\tan x = k\) also has \(\alpha + 180^\circ\). Add or subtract whole periods to stay inside the interval, and for \(\sin 2x\) widen the interval for \(2x\) first.
Fig. 1: sin x = 0.4 for 0° ≤ x ≤ 360°. The calculator gives 23.6°; the symmetry of the sine curve about x = 90° gives the second solution 180° − 23.6° = 156.4°
Section 01
Principal values.
In the lab, move the horizontal line \(y = k\) and watch where it cuts the curve. The calculator's answer is always the crossing inside one particular range, the principal value; the other crossings come from the shape of the graph.
\(\sin^{-1}x\), \(\cos^{-1}x\), \(\tan^{-1}x\) mean "the angle whose sine (cosine, tangent) is \(x\)". Each one is the inverse of the function restricted to a range where it is one-one.
\(\sin^{-1}x\) lies in \(-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}\) (\(-90^\circ\) to \(90^\circ\)).
\(\cos^{-1}x\) lies in \(0 \le y \le \pi\) (\(0^\circ\) to \(180^\circ\)).
\(\tan^{-1}x\) lies in \(-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}\) (\(-90^\circ\) to \(90^\circ\)).
Worked example Exact principal values
Find the exact values of (a) \(\sin^{-1}\left(-\tfrac{1}{2}\right)\) (b) \(\cos^{-1}\left(-\tfrac{1}{2}\right)\) (c) \(\tan^{-1}\left(-\sqrt{3}\right)\).
(a) \(\sin\dfrac{\pi}{6} = \tfrac{1}{2}\); the principal value must lie between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\), so take \(-\dfrac{\pi}{6}\).
(b) The range of \(\cos^{-1}\) is \(0\) to \(\pi\); a negative cosine needs the second quadrant: \(\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}\).
(c) \(\tan\dfrac{\pi}{3} = \sqrt{3}\); the principal value of \(\tan^{-1}\) of a negative number is negative.
\(\sin^{-1}x\) is NOT \(\dfrac{1}{\sin x}\). The \(-1\) means inverse function, just as \(f^{-1}\) does; \((\sin x)^{-1}\) is a different thing.
Section 02
Finding every solution.
Get the principal value, use the symmetry of the graph to get a second value, then add or subtract the period until you have every solution in the interval. A quick sketch of the graph and the line \(y = k\) tells you how many solutions to expect.
Equation
Second solution
Then add or subtract
\(\sin x = k\)
\(180^\circ - \alpha\) (\(\pi - \alpha\))
\(360^\circ\) (\(2\pi\))
\(\cos x = k\)
\(360^\circ - \alpha\) (\(2\pi - \alpha\)), or \(-\alpha\)
\(360^\circ\) (\(2\pi\))
\(\tan x = k\)
\(\alpha + 180^\circ\) (\(\alpha + \pi\))
\(180^\circ\) (\(\pi\))
Worked example Sine, in degrees
Solve \(\sin x = 0.4\) for \(0^\circ \le x \le 360^\circ\).
Principal value: \(x = \sin^{-1}0.4 = 23.578\ldots^\circ\).
Second solution by symmetry: \(180^\circ - 23.578^\circ = 156.42\ldots^\circ\).
Adding \(360^\circ\) to either takes us out of the interval, so there are just two (as Fig. 1 shows).
Answer: \(x = 23.6^\circ\) or \(156.4^\circ\)
Worked example Cosine, in radians
Solve \(\cos x = -0.3\) for \(0 \le x \le 2\pi\).
Principal value (radian mode): \(x = \cos^{-1}(-0.3) = 1.8755\).
Second solution: \(2\pi - 1.8755 = 4.4077\).
Answer: \(x = 1.88\) or \(4.41\)
Worked example Tangent, negative interval
Solve \(\tan x = -2\) for \(-180^\circ < x < 180^\circ\).
Principal value: \(\tan^{-1}(-2) = -63.43\ldots^\circ\), which is inside the interval.
The period of tan is \(180^\circ\): \(-63.43^\circ + 180^\circ = 116.57^\circ\), also inside; subtracting \(180^\circ\) goes outside.
Answer: \(x = -63.4^\circ\) or \(116.6^\circ\)
Examiner trap
A negative principal value is not automatically wrong, and it is not automatically an answer. If the interval is \(0^\circ\) to \(360^\circ\) and the calculator gives \(-63.4^\circ\), add \(180^\circ\) (or \(360^\circ\)) to bring it into range.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What is the principal value of \(\cos^{-1}\left(-\tfrac{1}{2}\right)\)?
Quick check+10 XP
One solution of \(\sin x = 0.6\) is \(36.9^\circ\). What is the other solution between \(0^\circ\) and \(360^\circ\)?
Quick check+10 XP
Solve \(\tan x = 1\) for \(0 \le x \le 2\pi\).
Quick check+10 XP
In which range does the principal value of \(\sin^{-1}x\) lie?
Section 03
Multiple angles and shifts.
For \(\sin 2x = k\), or \(\tan(x - 30^\circ) = k\), solve for the whole bracket first. Change the interval to match: if \(0 \le x \le \pi\), then \(0 \le 2x \le 2\pi\). Find every value of the bracket in the new interval, then convert back to \(x\) at the very end.
Worked example A double angle, exact
Solve \(2\sin 2x + 1 = 0\) for \(-\pi < x < \pi\), giving exact answers.
\(\sin\dfrac{\pi}{6} = \tfrac{1}{2}\), so the principal value is \(2x = -\dfrac{\pi}{6}\); by symmetry \(2x = -\pi + \dfrac{\pi}{6} = -\dfrac{5\pi}{6}\).
Add \(2\pi\) to each: \(2x = \dfrac{11\pi}{6}\) and \(\dfrac{7\pi}{6}\). Four values of \(2x\) in all.
Solve \(\tan(x - 30^\circ) = \sqrt{3}\) for \(0^\circ \le x \le 360^\circ\).
New interval: \(-30^\circ \le x - 30^\circ \le 330^\circ\).
\(\tan^{-1}\sqrt{3} = 60^\circ\); the period is \(180^\circ\), so \(x - 30^\circ = 60^\circ\) or \(240^\circ\).
Add \(30^\circ\) to each.
Answer: \(x = 90^\circ\) or \(270^\circ\)
Your turn
Solve \(\sin(x + 20^\circ) = 0.5\) for \(0^\circ \le x \le 360^\circ\).
\(20^\circ \le x + 20^\circ \le 380^\circ\). \(x + 20^\circ = 30^\circ\) or \(150^\circ\) (\(390^\circ\) is too big). So \(x = 10^\circ\) or \(130^\circ\).
Examiner trap
Halving too early loses solutions. If you solve \(\sin 2x = -\tfrac{1}{2}\) only for \(-\pi < 2x < \pi\) you find two answers instead of four.
Section 04
Equations that need an identity.
If an equation mixes \(\sin\) and \(\cos\), make it a single function. Two moves cover Paper 1: divide by \(\cos x\) to turn \(a\sin x = b\cos x\) into \(\tan x = \dfrac{b}{a}\), and use \(\sin^2 x + \cos^2 x = 1\) to turn a quadratic into one in a single function, then factorise.
Worked example Sine equals a multiple of cosine
Solve \(2\sin x = 5\cos x\) for \(0^\circ \le x \le 360^\circ\).
Divide by \(\cos x\) (it cannot be 0 here, or \(\sin x\) would be 0 too): \(\tan x = 2.5\).
Solve \(3\tan x = 2\cos x\) for \(0 \le x \le 2\pi\).
Multiply by \(\cos x\): \(3\sin x = 2\cos^2 x = 2 - 2\sin^2 x\), so \(2\sin^2 x + 3\sin x - 2 = 0\), \((2\sin x - 1)(\sin x + 2) = 0\). \(\sin x = \tfrac{1}{2}\), so \(x = \dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\).
Examiner trap
Never cancel a trig function that could be zero. In \(\sin x = 2\sin x\cos x\), dividing by \(\sin x\) loses the solutions of \(\sin x = 0\). Instead factorise: \(\sin x(1 - 2\cos x) = 0\).
Stage 2 · Exam
Exam-style questions
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Solve \(2\cos 2x + 1 = 0\) for \(0 \le x \le \pi\).
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Solve \(3\tan x = 4\sin x\) for \(0^\circ \le x \le 180^\circ\).
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Solve \(2\sin^2\theta = 3\cos\theta\) for \(0^\circ \le \theta \le 360^\circ\).
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Solve \(\sin\left(x - \dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}\) for \(0 \le x \le 2\pi\).
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