One value on the calculator, every value in the domain.
Your calculator gives one answer to \(\sin x = 0.6\), but the graph crosses the line \(y = 0.6\) again and again. Learn to find the principal value, use symmetry to collect every other solution in the domain, handle multiple and compound angles, and use the identities to turn harder equations into one function you can solve.
Make one trigonometric function the subject, find the principal value on your calculator, then use symmetry to list every solution in the domain: if \(\alpha\) is one solution, the others in one turn are \(180^\circ - \alpha\) for sine, \(360^\circ - \alpha\) for cosine and \(180^\circ + \alpha\) for tangent (or \(\pi - \alpha\), \(2\pi - \alpha\), \(\pi + \alpha\) in radians), and you add or subtract whole periods to stay in the domain. Change \(\sec\), \(\operatorname{cosec}\) and \(\cot\) to \(\cos\), \(\sin\) and \(\tan\) first, adjust the domain before solving in a multiple angle such as \(2x\), and factorise rather than divide by a function that could be zero.
Fig. 1: the graph of y = sin x for 0° ≤ x ≤ 360° and the line y = 0.6. They meet twice: the calculator gives 36.9°, and symmetry about x = 90° gives the second solution 180° - 36.9° = 143.1°
Section 01
The principal value, then the rest.
Rearrange to \(\sin x = k\), \(\cos x = k\) or \(\tan x = k\). The inverse key on your calculator gives the principal value \(\alpha\): between \(-90^\circ\) and \(90^\circ\) for \(\sin^{-1}\) and \(\tan^{-1}\), between \(0^\circ\) and \(180^\circ\) for \(\cos^{-1}\). The graph is symmetric, so a second solution in each turn comes from the rule below. Then add or subtract \(360^\circ\) (or \(2\pi\)) until every solution in the domain is listed. If \(|k| > 1\) for sine or cosine, there are no solutions.
Equation
Solutions in one turn (degrees)
In radians
Then repeat every
\(\sin x = k\)
\(\alpha\) and \(180^\circ - \alpha\)
\(\alpha\) and \(\pi - \alpha\)
\(360^\circ\) or \(2\pi\)
\(\cos x = k\)
\(\alpha\) and \(360^\circ - \alpha\)
\(\alpha\) and \(2\pi - \alpha\)
\(360^\circ\) or \(2\pi\)
\(\tan x = k\)
\(\alpha\) and \(180^\circ + \alpha\)
\(\alpha\) and \(\pi + \alpha\)
\(180^\circ\) or \(\pi\)
The CAST diagram says the same thing: sine is positive in the first and second quadrants, tangent in the first and third, cosine in the first and fourth. For the reciprocal functions, turn them over first: \(\sec x = 2\) means \(\cos x = \frac{1}{2}\), \(\operatorname{cosec} x = k\) means \(\sin x = \frac{1}{k}\), \(\cot x = k\) means \(\tan x = \frac{1}{k}\). Give answers in degrees to 1 decimal place, and in radians to 3 significant figures or as exact multiples of \(\pi\).
Worked example Sine: use the symmetry about 90 degrees
Solve \(5\sin x = 3\) for \(0^\circ \le x \le 360^\circ\).
\(\sin x = 0.6\), so the principal value is \(\alpha = \sin^{-1} 0.6 = 36.87^\circ\).
Sine is also positive in the second quadrant: \(180^\circ - 36.87^\circ = 143.13^\circ\).
Adding \(360^\circ\) to either goes past \(360^\circ\), so there are no more (Fig. 1).
Answer: \(x = 36.9^\circ\) or \(x = 143.1^\circ\)
Worked example A reciprocal function
Solve \(2\sec x + 5 = 0\) for \(0^\circ \le x \le 360^\circ\).
\(\sec x = -2.5\), so \(\cos x = \dfrac{1}{-2.5} = -0.4\).
Principal value: \(\cos^{-1}(-0.4) = 113.58^\circ\).
Cosine: the other solution is \(360^\circ - 113.58^\circ = 246.42^\circ\).
Answer: \(x = 113.6^\circ\) or \(x = 246.4^\circ\)
Your turn
Solve \(\operatorname{cosec} x = 4\) for \(0^\circ \le x \le 360^\circ\).
\(\sin x = \frac{1}{4}\), so \(x = 14.5^\circ\) or \(x = 180^\circ - 14.5^\circ = 165.5^\circ\).
Examiner trap
The calculator's value may be outside the domain. For \(\tan x = -2\) it gives \(-63.4^\circ\); that is not an answer in \(0^\circ \le x \le 360^\circ\). Add \(180^\circ\) to get \(116.6^\circ\) and again to get \(296.6^\circ\). Also, \(\sec x = 2\) is \(\cos x = \frac{1}{2}\), never \(\cos x = 2\).
Section 02
Multiple and compound angles.
For an equation in \(2x\), \(\frac{x}{3}\) or \(2x - \frac{\pi}{4}\), call the whole bracket \(u\). Change the domain for \(x\) into a domain for \(u\) first, solve for \(u\) in that wider (or narrower) domain, and only then turn each \(u\) back into \(x\).
Step 1: if \(0^\circ \le x \le 360^\circ\), then \(0^\circ \le 2x \le 720^\circ\), so look for solutions of \(\sin 2x = k\) over two full turns.
Step 2: list every value of \(u\) in the new domain using the symmetry rules.
Step 3: undo the bracket for each value: divide by 2, multiply by 3, add \(\frac{\pi}{4}\) then divide by 2, and so on.
Worked example A double angle in degrees
Solve \(\sin 2x = -0.4\) for \(0^\circ \le x \le 360^\circ\).
Let \(u = 2x\). The domain becomes \(0^\circ \le u \le 720^\circ\).
Principal value \(\sin^{-1}(-0.4) = -23.58^\circ\), outside the domain. The values in one turn are \(180^\circ + 23.58^\circ = 203.58^\circ\) and \(360^\circ - 23.58^\circ = 336.42^\circ\).
Add \(360^\circ\) for the second turn: \(563.58^\circ\) and \(696.42^\circ\).
Halve all four values.
Answer: \(x = 101.8^\circ,\ 168.2^\circ,\ 281.8^\circ\) or \(348.2^\circ\)
Worked example A compound angle in radians, exact answers
Solve \(\sec\left(2x - \dfrac{\pi}{4}\right) = 2\) for \(0 \le x \le \pi\).
\(0 \le x \le \pi\) gives \(-\dfrac{\pi}{4} \le u \le \dfrac{7\pi}{4}\).
\(\cos u = \dfrac{1}{2}\): \(u = \dfrac{\pi}{3}\) or \(u = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}\). (The value \(-\dfrac{\pi}{3}\) is below \(-\dfrac{\pi}{4}\), so it is not in the domain.)
\(2x = u + \dfrac{\pi}{4}\): \(2x = \dfrac{7\pi}{12}\) or \(2x = \dfrac{23\pi}{12}\).
Answer: \(x = \dfrac{7\pi}{24}\) or \(x = \dfrac{23\pi}{24}\)
Your turn
Solve \(2\cos\dfrac{x}{3} = -1\) for \(0^\circ \le x \le 720^\circ\).
Let \(u = \frac{x}{3}\), so \(0^\circ \le u \le 240^\circ\). \(\cos u = -\frac{1}{2}\) gives \(u = 120^\circ\) or \(240^\circ\), so \(x = 360^\circ\) or \(x = 720^\circ\).
Examiner trap
Do not solve \(\sin 2x = -0.4\) over \(0^\circ\) to \(360^\circ\) and then halve: that gives only two of the four answers. Widen the domain for \(2x\) FIRST, and halve at the very end.
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
Solve \(\tan x = -1\) for \(0^\circ \le x \le 360^\circ\).
Quick check+10 XP
Solve \(\sec x = 2\) for \(0^\circ \le x \le 360^\circ\).
Quick check+10 XP
To solve \(\sin 2x = 0.5\) for \(0^\circ \le x \le 360^\circ\), over which interval should you find values of \(2x\)?
Quick check+10 XP
Solve \(2\sin^{2} x + \sin x - 1 = 0\) for \(0^\circ \le x \le 360^\circ\).
Section 03
Use an identity to get one function.
When an equation mixes functions, make it one function before you solve. Three tools do almost every job: \(\dfrac{\sin x}{\cos x} = \tan x\), \(\sin^{2} x + \cos^{2} x = 1\), and its two relatives \(\sec^{2} x = 1 + \tan^{2} x\) and \(\operatorname{cosec}^{2} x = 1 + \cot^{2} x\). A quadratic in one function is then factorised like any quadratic.
\(a\sin x = b\cos x\): divide by \(\cos x\) (it cannot be zero here, since then \(\sin x\) would be zero too) to get \(\tan x = \frac{b}{a}\).
\(\cos^{2} x\) with \(\sin x\): replace \(\cos^{2} x\) by \(1 - \sin^{2} x\). \(\sec^{2} x\) with \(\tan x\): replace \(\sec^{2} x\) by \(1 + \tan^{2} x\).
Reject impossible roots: \(\sin x\) and \(\cos x\) lie between \(-1\) and \(1\); \(\sec x\) and \(\operatorname{cosec} x\) are never between \(-1\) and \(1\).
Never divide by a function that could be zero. In \(2\sin x\cos x = \cos x\), take out the factor: \(\cos x(2\sin x - 1) = 0\), so \(\cos x = 0\) or \(\sin x = \frac{1}{2}\).
Worked example A quadratic in sin x
Solve \(2\cos^{2} x + 3\sin x = 3\) for \(0^\circ \le x \le 360^\circ\).
Replace \(\cos^{2} x\) by \(1 - \sin^{2} x\): \(2 - 2\sin^{2} x + 3\sin x = 3\).
Rearrange: \(2\sin^{2} x - 3\sin x + 1 = 0\), so \((2\sin x - 1)(\sin x - 1) = 0\).
\(\sin x = \dfrac{1}{2}\): \(x = 30^\circ\) or \(150^\circ\).
\(\sin x = 1\): \(x = 90^\circ\).
Answer: \(x = 30^\circ,\ 90^\circ\) or \(150^\circ\)
Worked example Using sec squared = 1 + tan squared
Solve \(3\sec^{2} x = 5\tan x + 5\) for \(0^\circ \le x \le 360^\circ\).
\(3(1 + \tan^{2} x) = 5\tan x + 5\), so \(3\tan^{2} x - 5\tan x - 2 = 0\).
Factorise: \((3\tan x + 1)(\tan x - 2) = 0\).
\(\tan x = 2\): \(\alpha = 63.43^\circ\), and \(63.43^\circ + 180^\circ = 243.43^\circ\).
\(\tan x = -\dfrac{1}{3}\): \(\tan^{-1}\left(-\dfrac{1}{3}\right) = -18.43^\circ\); add \(180^\circ\) and \(360^\circ\) to get \(161.57^\circ\) and \(341.57^\circ\).
Answer: \(x = 63.4^\circ,\ 161.6^\circ,\ 243.4^\circ\) or \(341.6^\circ\)
Your turn
Solve \(2\sin^{2} x - 5\sin x - 3 = 0\) for \(0^\circ \le x \le 360^\circ\).
\((2\sin x + 1)(\sin x - 3) = 0\). \(\sin x = 3\) is impossible, so \(\sin x = -\frac{1}{2}\), giving \(x = 210^\circ\) or \(330^\circ\).
Examiner trap
Cancelling \(\cos x\) from both sides of \(2\sin x\cos x = \cos x\) leaves \(\sin x = \frac{1}{2}\) and throws away \(x = 90^\circ\) and \(270^\circ\), where \(\cos x = 0\). Factor out the common function instead, and solve each factor.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to IGCSE Additional Mathematics Paper 1 and Paper 2 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
Solve \(2\sin x\cos x = \cos x\) for \(0^\circ \le x \le 360^\circ\).
Exam style+20 XP
Solve \(2\cos\dfrac{x}{2} = \sqrt{3}\) for \(0 \le x \le 4\pi\), giving exact answers.
Exam style+20 XP
Solve \(\cot x = 2\tan x\) for \(0^\circ < x < 180^\circ\).
Exam style+20 XP
Solve \(2\cos^{2} x = 3\sin x\) for \(0^\circ \le x \le 360^\circ\).
Skill unlocked
Every solution in the domain, found and checked.
This skill is now recorded on your profile. Keep the chain going.