Solve a trig equation on an interval

In 7 of 20 recent exams
1% of the marks
Paper 1 or 2
2 to 6 marks

How it is set in the exam

Paper 1 (no calculator), 2–6 marks a part; common (7 of 20 exam versions, all on Paper 1 (no calculator)), in 2 of the 7 as "hence solve" after a "show that" identity. The same solving is the last step of many calculus parts (Where it hides). On Paper 2 (calculator) an equation that mixes trig with other terms is solved graphically: Solve graphically, with zeros and intersections; one ratio equal to a constant with no exact angle still follows the Steps, with $\sin^{-1}$ (or $\cos^{-1}$ , $\tan^{-1}$ ) for the reference angle (angle from a ratio).

Spot it in the exam

Exam phrases to spot (5)

“solve … for $0 \le x \le 2\pi$ ”
“find all the values of $\theta$ ”
“the least positive value of $x$ ”
“hence solve”
“show that $\tan\theta = \ldots$ ”

In disguise (2)

a condition on a solid or a figure (two areas, or two lengths, equal) that becomes $\tan\theta = k$ once a common $\sin\theta$ is taken out and $\sin\theta = 0$ is rejected (a length or an angle can't be $0$ ); in a solid, the lengths come from Triangles inside a solid
an equation built from a composite, such as $f(2x) = \ldots$ , where forming the composite is only the set-up

Variants (6)

a compound argument ( $2x$ , $\frac{x}{2} + c$ , $3x + 15°$ ), in radians or degrees; the least positive solution
$\sin$ and $\cos$ of the same argument → $\tan$ , on the doubled interval
squares of $\sin$ and $\cos$ → $\pm$
an identity first ("show that"), then factorise into two ratios, keeping the $\sin x = 0$ branch
a common factor $\sin\theta$ rejected in context, ending in $\tan\theta = k$
an identity turns the equation into $\sin 2\theta = k$ on an interval in degrees

Looks like it, but isn't (4)

a model in a context equal to a level ("when is the height $6$ m", "for how long is it above"): Above or below a level
a quadratic in one ratio (three terms with a constant, powers $2 : 1$ ): Quadratic in disguise
one ratio of an unknown angle is given, or a triangle gives it (the cosine rule; the sine rule with angles $\theta$ and $2\theta$ , even though $\sin\theta$ is cancelled on the way), and another ratio is asked ("find the exact value of $\sin 2\theta$ "): Exact values without the angle: from one ratio to another
trig mixed with $x$ outside it ( $x = 2\sin x$ ), on : Solve graphically. One ratio equal to a value with no exact angle stays here: $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ gives the reference angle.

Worked example

A compound argument: the least positive solution

Paper 1 (no calculator)

Question

Find the least positive value of $x$ for which $2\sin\left(2x + \frac{5\pi}{6}\right) = \sqrt{3}$ . [5]

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Which method?

one ratio of one argument $=$ a constant
First move
reference angle, then the quadrants; for a compound argument, change the interval first
Example
$2\sin\left(2x + \frac{5\pi}{6}\right) = \sqrt{3}$
$\sin$ and $\cos$ of the same argument, one term each, and no constant
First move
divide by $\cos$ : $\tan(\text{argument}) = k$
Example
$\sin 2x + \sqrt{3}\cos 2x = 0$
squares of $\sin$ and $\cos$ , with or without a constant
First move
no constant: divide by $\cos^2$ to get $\tan^2$ ; with a constant: $\cos^2 = 1 - \sin^2$ to leave one square (or first write the constant $k$ as $k(\sin^2 + \cos^2)$ , then divide by $\cos^2$ , after checking that $\cos = 0$ is not a solution); then $\pm$
Example
$3\sin^2\theta = 1 + \cos^2\theta$
a common factor, such as $\sin x$ on both sides
First move
bring everything to one side and factorise; never cancel in a bare equation
Example
$\sin 2x\cos x = \sin x$
a double angle next to single angles, after a "show that"
First move
use the identity from the earlier part, then one of the rows above
Example
Example 3.3
a quadratic in one ratio
First move
Quadratic in disguise
Example
$2\cos^2 x - 3\cos x - 2 = 0$
one ratio $=$ a value with no exact reference angle ()
First move
the reference angle from $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ (GDC: angle from a ratio), then the quadrants; a graph checks the count
Example
$3\sin 2x = 1$
trig with non-trig terms ()
First move
Solve graphically, with zeros and intersections
Example
$3\cos x = x + 1$

The equation	First move	Example
one ratio of one argument $=$ a constant	reference angle, then the quadrants; for a compound argument, change the interval first	$2\sin\left(2x + \frac{5\pi}{6}\right) = \sqrt{3}$
$\sin$ and $\cos$ of the same argument, one term each, and no constant	divide by $\cos$ : $\tan(\text{argument}) = k$	$\sin 2x + \sqrt{3}\cos 2x = 0$
squares of $\sin$ and $\cos$ , with or without a constant	no constant: divide by $\cos^2$ to get $\tan^2$ ; with a constant: $\cos^2 = 1 - \sin^2$ to leave one square (or first write the constant $k$ as $k(\sin^2 + \cos^2)$ , then divide by $\cos^2$ , after checking that $\cos = 0$ is not a solution); then $\pm$	$3\sin^2\theta = 1 + \cos^2\theta$
a common factor, such as $\sin x$ on both sides	bring everything to one side and factorise; never cancel in a bare equation	$\sin 2x\cos x = \sin x$
a double angle next to single angles, after a "show that"	use the identity from the earlier part, then one of the rows above	Example 3.3
a quadratic in one ratio	Quadratic in disguise	$2\cos^2 x - 3\cos x - 2 = 0$
one ratio $=$ a value with no exact reference angle ()	the reference angle from $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ (GDC: angle from a ratio), then the quadrants; a graph checks the count	$3\sin 2x = 1$
trig with non-trig terms ()	Solve graphically, with zeros and intersections	$3\cos x = x + 1$

Steps

Rewrite
until each piece is one ratio of one argument equal to a constant: use the identity from a "show that"; divide $\sin$ by $\cos$ to get $\tan$ ; take square roots with $\pm$ ; factorise a common factor (cancel it only where the question rules its zero out, and say why).
Reject impossible values
with a reason: $\lvert \sin \rvert \le 1$ and $\lvert \cos \rvert \le 1$ .
Change the interval
for a compound argument: if $u = bx + c$ , do the same to both ends of the interval.
Reference angle
from the exact values (on , from $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ if the value is not a standard one), then every $u$ in the new interval (the table). A sketch of the graph on the interval confirms the count.
Undo the argument:
solve each $u = bx + c$ for $x$ .
Answer:
only the values in the interval, in its unit (radians unless degrees are shown), with every rejected value gone.

Common mistakes

Dividing by a common factor
such as $\sin x$ in a bare equation: its solutions are lost, so factorise instead (Example 3.3). Where the question rules the factor's zero out (an angle of a triangle or a solid, $0 < \theta < \frac{\pi}{2}$ ), cancelling is allowed, but the line " $\sin\theta \ne 0$ because …" is itself a mark ().
Only the calculator's angle,
or only the first quadrant: $\sin x = \frac{1}{2}$ has two solutions in every turn.
Not changing the interval
for $2x$ or $2x + c$ , and stopping after one turn. Missing or extra solutions lose the final ; extras are condoned only at the $2\theta$ stage, never in the answer.

More mistakes (7)

Values outside the interval
left in the answer, or a rejected value still there: the final is lost.
The first angle in a "least positive" question
the first value of the argument can give $x < 0$ (Example 3.1). Check the sign of $x$ , reject it with its reason (the ), and give one value only.
The wrong unit
degree answers to an interval in radians (or the reverse) lose a mark even when the angles are right.
The $\pm$ forgotten
after a square root: accuracy marks are lost.
Setting the argument equal to the value
for $\cos(3x + 15°) = \frac{1}{2}$ , writing $3x + 15° = \frac{1}{2}$ earns no method mark.
The "show that" before "hence solve"
start from one side and write each identity as you use it. A jump from the $\frac{\sin}{\cos}$ substitution straight to the result loses the last mark (Showing an identity).
No working on
correct answers alone have been capped at 3 marks out of 5.

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