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You solve trigonometric equations in a given interval, in degrees. These include equations with multiple angles such as \(\cos 2x\), shifted angles such as \(\sin(x + 30^\circ)\), and quadratics in sin, cos or tan. Many questions need an identity first, and you must find every solution in the interval.
Key facts
\(\sin x = k\): \(x = \alpha\) or \(180^\circ - \alpha\), then add multiples of 360°
\(\cos x = k\): \(x = \alpha\) or \(360^\circ - \alpha\), then add multiples of 360°
\(\tan x = k\):\(x = \alpha\), then add multiples of 180°
For \(\sin bx\): multiply the interval by b, solve, then divide by b
For \(\sin(x + c)\): add c to both ends of the interval, solve, then subtract c
\(|\sin x| \le 1\) and \(|\cos x| \le 1\); tan x can take any value
Notes
The basic method
diagram
Find the principal value with \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\) on your calculator.
Use symmetry to find the rest: if \(\sin x = k\), x = α or 180° − α; if \(\cos x = k\), x = α or 360° − α; if \(\tan x = k\), x = α or α + 180°. Then add or subtract 360° (180° for tan) as needed.
sin x = 0.5 has two solutions from 0° to 360°: α = 30° and 180° − α = 150°.
The principal value may be outside the interval (e.g. negative); you still use it to find the solutions inside.
A quick sketch of the graph with the line y = k shows how many solutions to expect.
Multiple and shifted angles
For \(\sin 3x = k\) with \(0 \le x \lt 360^\circ\): let u = 3x, so \(0 \le u \lt 1080^\circ\). Find every u, then divide by 3.
For \(\cos(x - 40^\circ) = k\) with \(0 \le x \lt 360^\circ\): let \(u = x - 40^\circ\), so \(-40^\circ \le u \lt 320^\circ\). Find every u, then add 40°.
Always change the interval first, solve for u, then convert back to x.
Quadratics and identities
Treat the trig function as the unknown: \(3\tan^2 x - 5\tan x - 2 = 0\) factorises to \((3\tan x + 1)(\tan x - 2) = 0\)
If there is a mix of functions, use an identity to get just one: \(\sin^2 x + \cos^2 x = 1\),\(1 + \tan^2 x = \sec^2 x\), or a double angle formula such as \(\cos 2x = 1 - 2\sin^2 x\)
Reject values outside the range of the function: \(\sin x = -2\) has no solutions.
Don't lose solutions
Dividing \(\sin x = 2\cos x\) by \(\cos x\) is safe: if \(\cos x = 0\) then \(\sin x = \pm 1\), which does not satisfy the equation.
If you square both sides, check each answer in the original equation, because squaring can create extra solutions.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Solve an equation with a multiple angle
4 marksGrade C
Replace the multiple angle with u and multiply the interval.
Find the principal value and use symmetry for the second value.
Add 360° to get every value of u in the new interval.
Divide by the multiple and round at the end.
Example. Solve, for \(0 \le x \lt 360^\circ\), the equation \(\cos 2x = -0.3\), giving your answers to 1 decimal place.
Adding 360°: \(u = 467.46^\circ\) or \(612.54^\circ\)
M1
x = 53.7°, 126.3°, 233.7°, 306.3°
A1
Solve a quadratic in tan x
5 marksGrade B
Factorise as a quadratic in tan x.
Solve each factor separately.
For tan, add 180° to find the next solution.
Example. Solve, for \(0 \le x \lt 360^\circ\), the equation \(3\tan^2 x - 5\tan x - 2 = 0\), giving your answers to 1 decimal place.
Show the model answerHide the model answer
\((3\tan x + 1)(\tan x - 2) = 0\)
M1
\(\tan x = 2\) or \(\tan x = -\dfrac{1}{3}\)
A1
\(\tan x = 2\): x = 63.4°, 243.4°
M1
\(\tan x = -\dfrac{1}{3}\): the principal value is −18.4°, so x = 161.6°, 341.6°
A1 A1
Use an identity, then solve
5 marksGrade A
Choose the form of \(\cos 2x\) that matches the other terms (here \(1 - 2\sin^2 x\)).
Rearrange into a quadratic equal to zero and factorise.
Reject any impossible value, then solve the rest.
Example. Solve, for \(0 \le x \lt 360^\circ\), the equation \(\cos 2x = 3\sin x - 1\)
Show the model answerHide the model answer
\(1 - 2\sin^2 x = 3\sin x - 1\)
M1
\(2\sin^2 x + 3\sin x - 2 = 0\)
A1
\((2\sin x - 1)(\sin x + 2) = 0\)
M1
\(\sin x = -2\) has no solutions, so \(\sin x = \dfrac{1}{2}\)
A1
x = 30°, 150°
A1
Don’t lose marks
Giving only the calculator value and missing the other solutions.
Dividing by 2 (for \(\sin 2x\)) before finding all the solutions for 2x.
Dividing both sides by sin x or cos x and losing solutions.
Including values outside the interval, e.g. 360° when the interval is \(0 \le x \lt 360^\circ\)
More tips
Memory tricks
Sketch the graph over the (adjusted) interval and draw the line y = k: the number of crossings is the number of solutions.
For tan, just keep adding 180° (or π).
Check each answer by substituting it back into the original equation on your calculator.
For \(\sin kx = c\) or \(\cos kx = c\) with \(-1 \lt c \lt 1\), there are usually 2k solutions in a 360° interval.
Exam technique
Unless told otherwise, give angles in degrees to 1 d.p. and in radians to 3 s.f., or as exact multiples of π when the question asks for exact answers.
Read the interval carefully: \(\le\) or \(\lt\), degrees or radians.
If the question says 'Solutions relying entirely on calculator technology are not acceptable', show the algebra and each step to the angles.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Solve \(\sin x=k\) in a given intervalFind the calculator value, then use symmetry (or CAST) to find every other solution in the interval.
Grade D
Use \(\tan x=\dfrac{\sin x}{\cos x}\) to solve equationsTurn \(3\sin x = 2\cos x\) into \(\tan x = \dfrac{2}{3}\) and solve.
Grade C
Solve equations with multiple anglesFor \(\cos 2x = k\), solve for 2x in the doubled interval, then halve.
Grade C
Solve equations with a shifted angleFor \(\cos(x - \alpha) = k\), shift the interval by α, solve, then add α back on.
Grade B
Solve quadratic equations in sin, cos or tanFactorise, solve each factor, and reject any impossible value.
Grade A
Use identities before solvingUse a Pythagorean or double angle identity to get an equation in one trig function.
Grade A*
Factorise instead of dividing to keep solutionse.g. \(\sin 2x = \sin x\) gives \(\sin x(2\cos x - 1) = 0\), so the solutions of \(\sin x = 0\) are not lost.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
A student is asked to solve the equation \(2\sin x\cos x = \sin x\) for \(0 \le x \lt 360^\circ\). The student writes: Dividing both sides by \(\sin x\): \(2\cos x = 1\), so \(\cos x = \frac{1}{2}\) and \(x = 60^\circ\) or \(x = 300^\circ\).
(a) Explain the error in the student’s method.[1]
(b) Find all the solutions of the equation for \(0 \le x \lt 360^\circ\).[2]
Show the answer and mark scheme
(a)Answer: Dividing by \(\sin x\) loses the solutions where \(\sin x = 0\).
B1 for dividing by \(\sin x\) loses the solutions with \(\sin x = 0\) (you cannot divide by zero)
Worked solution: The student divided by \(\sin x\), which is not allowed when \(\sin x = 0\). The solutions with \(\sin x = 0\) have been lost.
M1 for a correct principal value, \(3x + 15^\circ = 30^\circ\)
A1 for one correct value of \(x\)
M1 for a correct method to find further values of \(3x + 15^\circ\) in the range \(-525^\circ \le 3x + 15^\circ \lt 555^\circ\)
A1 for at least 5 correct values of \(x\)
A1 for all 6 values and no others in the interval
Worked solution: \(\cos\left(3x + 15^\circ\right) = \frac{\sqrt{3}}{2}\). Let \(u = 3x + 15^\circ\). For \(-180^\circ \le x \lt 180^\circ\), \(-525^\circ \le u \lt 555^\circ\). Principal value \(u = 30^\circ\); the values of \(u\) in the range are \(u = -390^\circ, \ -330^\circ, \ -30^\circ, \ 30^\circ, \ 330^\circ, \ 390^\circ\). So \(x = -135^\circ, \ -115^\circ, \ -15^\circ, \ 5^\circ, \ 105^\circ, \ 125^\circ\).
Question 3Hard9 marks
(a) Solve, for \(0 \le \theta \lt 360^\circ\), the equation \[2\tan^2\theta + 7\tan\theta + 6 = 0.\] Give your answers to 1 decimal place.[5]
(b) Hence solve, for \(0 \le x \lt 180^\circ\), the equation \[2\tan^2(2x - 40^\circ) + 7\tan(2x - 40^\circ) + 6 = 0.\] Give your answers to 1 decimal place.[4]
M1 for setting \(2x - 40^\circ\) equal to their values from part (a), using the range \(-40^\circ \le 2x - 40^\circ \lt 320^\circ\)
A1 for one correct value of \(x\)
dM1 for a correct method to find all values of \(2x - 40^\circ\) in the range (adding or subtracting \(180^\circ\) as needed)
A1 for all of \(78.3^\circ, \ 81.8^\circ, \ 168.3^\circ, \ 171.8^\circ\) and no others
Worked solution: Let \(\theta = 2x - 40^\circ\). For \(0 \le x \lt 180^\circ\), \(-40^\circ \le \theta \lt 320^\circ\). The solutions of part (a), with multiples of \(180^\circ\) added or subtracted, give \(\theta = 116.56\ldots^\circ, \ 123.69\ldots^\circ, \ 296.56\ldots^\circ, \ 303.69\ldots^\circ\) in this range. So \(x = \frac{\theta + 40^\circ}{2} = 78.3^\circ, \ 81.8^\circ, \ 168.3^\circ, \ 171.8^\circ\).