Harder Trig Equations

tan mixed with sin/cos: replace tan by sin/cos, multiply through by cos x (cos x ≠ 0), use sin² = 1 − cos², solve the quadratic

In the exam for Edexcel International AS Maths · 8 real past-paper questions below

tan with sin or cos

You can already solve a quadratic in $sin x$ or $cos x$ . Pure 2 adds a twist: tan in the same equation as sin or cos, such as $sin x tan x = 2$ . This type comes up often on Pure 2 papers.

The plan: get rid of tan, clear the fraction, and swap $sin^{2} x$ so only $cos x$ is left.

Steps

1
1. Split tan: write $tan x = \frac{s i n x}{c o s x}$ .
2. Clear: multiply every term by $cos x$ .
3. Swap: replace $sin^{2} x$ by $1 - cos^{2} x$ .
4. Solve the quadratic in $cos x$ , then find every angle in the range.

Worked example

1
Solve $sin x tan x + 1 = 5 cos x$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Show all stages of your working.
2
$sin x \cdot \frac{s i n x}{c o s x} + 1 = 5 cos x$ — Split tan into sin over cos.
3
$sin^{2} x + cos x = 5 cos^{2} x$ — Multiply every term by $cos x$ . The 1 becomes $cos x$ .
4
$1 - cos^{2} x + cos x = 5 cos^{2} x$ — Swap $sin^{2} x$ for $1 - cos^{2} x$ . Now only cos is left.
5
$6 cos^{2} x - cos x - 1 = 0$ — Collect everything on one side.
6
$(2 cos x - 1) (3 cos x + 1) = 0$ — Factorise, as if $cos x$ were a letter $c$ .
7
$cos x = \frac{1}{2} \Rightarrow x = 6 0^{\circ}, 30 0^{\circ}$ — Calculator value, then $36 0^{\circ} - 6 0^{\circ}$ .
8
$cos x = - \frac{1}{3} \Rightarrow x = 109. 5^{\circ}, 250. 5^{\circ}$ — $109.4 7^{\circ}$ , then $36 0^{\circ} - 109.4 7^{\circ}$ .

Rule

Why is multiplying by $cos x$ safe? Where $cos x = 0$ , tan does not exist, so those angles were never allowed. Multiplying cannot add a false answer.

Memory trick

Split, Clear, Swap. Split tan, clear the fraction with cos, swap sin squared. Then it is a quadratic you already know.

Rule

How the marks go: method marks for replacing tan and reaching a quadratic in $cos x$ only, an accuracy mark for the right quadratic, a method mark for solving it, then marks for the angles. Show the roots before the angles.

Common mistake

Forgetting to multiply the number term by $cos x$ . In the example, $1$ must become $cos x$ . Leaving it as 1 gives a wrong quadratic.

Key idea

Tan mixed with sin or cos? Split, clear, swap. It ends as a quadratic in one ratio.

Roots that cannot work

The quadratic often has one root that is impossible. Sin and cos are never bigger than 1 or smaller than −1. So a root like $cos x = \frac{3}{2}$ gives no angle.

Rule

Reject a root outside −1 to 1 and say why, for example: $cos x = \frac{3}{2}$ has no solutions, since $- 1 \leq cos x \leq 1$ .

Worked example

1
Solve $6 sin x tan x + 5 = 0$ for $0 \leq x < 2 π$ , in radians to 3 significant figures.
2
$6 sin^{2} x + 5 cos x = 0$ — Split tan, then multiply every term by $cos x$ .
3
$6 - 6 cos^{2} x + 5 cos x = 0$ — Swap $sin^{2} x$ for $1 - cos^{2} x$ , and expand.
4
$6 cos^{2} x - 5 cos x - 6 = 0$ — Multiply by −1 so the squared term is positive.
5
$(2 cos x - 3) (3 cos x + 2) = 0$ — Factorise.
6
$cos x = \frac{3}{2}$ — Reject: cos is never more than 1.
7
$cos x = - \frac{2}{3} \Rightarrow x = 2.30, 3.98$ — Calculator in radians gives 2.3005. The partner is $2 π - 2.3005 = 3.9827$ .

The same plan works when the angle is a compound angle. Call the inside angle $A$ , and change the range first.

Worked example

1
Solve $8 tan (x + 2 0^{\circ}) = 3 cos (x + 2 0^{\circ})$ for $0^{\circ} \leq x < 36 0^{\circ}$ .
2
$2 0^{\circ} \leq A < 38 0^{\circ}$ — Let $A = x + 2 0^{\circ}$ . Add 20° to both ends of the range.
3
$8 sin A = 3 cos^{2} A$ — Split tan, then multiply both sides by $cos A$ .
4
$8 sin A = 3 - 3 sin^{2} A$ — This time sin is left alone, so swap $cos^{2} A$ for $1 - sin^{2} A$ .
5
$(3 sin A - 1) (sin A + 3) = 0$ — Collect to $3 sin^{2} A + 8 sin A - 3 = 0$ and factorise.
6
$sin A = - 3$ — Reject: sin is never less than −1.
7
$sin A = \frac{1}{3} \Rightarrow A = 160. 5^{\circ}, 379. 5^{\circ}$ — $19.4 7^{\circ}$ is below 20°, so use $18 0^{\circ} - 19.4 7^{\circ}$ and $19.4 7^{\circ} + 36 0^{\circ}$ .
8
$x = 140. 5^{\circ}, 359. 5^{\circ}$ — Take 20° from each.

Common mistake

Swapping the wrong square. Swap whichever square leaves one ratio. If cos is the one standing alone, swap $sin^{2}$ . If sin is alone, swap $cos^{2}$ .

Key idea

A root above 1 or below −1 is not a mistake in your algebra. Cross it out, write why, and carry on with the other root.

Squares: roots and brackets

Two more AS types have a square in them. A squared ratio, like $cos^{2} (2 x + 2 0^{\circ}) = k$ , and a squared bracket, like $(sin x + 2 cos x)^{2} = 1$ . They need different first moves.

Rule

Squared ratio: square root both sides and keep both signs: a positive number has a + and a − square root.

Worked example

1
Solve $5 cos^{2} (2 x + 2 0^{\circ}) = 2$ for $0^{\circ} \leq x < 18 0^{\circ}$ .
2
$2 0^{\circ} \leq 2 x + 2 0^{\circ} < 38 0^{\circ}$ — Change the range: double, then add 20°.
3
$cos (2 x + 2 0^{\circ}) = \pm \frac{2}{5} = \pm 0.6325$ — Divide by 5, then square root. Keep both signs.
4
$+ 0.6325 : 2 x + 2 0^{\circ} = 50.7 7^{\circ}, 309.2 3^{\circ}$ — Calculator value and $36 0^{\circ} - 50.7 7^{\circ}$ .
5
$- 0.6325 : 2 x + 2 0^{\circ} = 129.2 3^{\circ}, 230.7 7^{\circ}$ — Calculator value and $36 0^{\circ} - 129.2 3^{\circ}$ .
6
$x = 15. 4^{\circ}, 54. 6^{\circ}, 105. 4^{\circ}, 144. 6^{\circ}$ — Take 20° from each and halve. Four answers in all.

Rule

Squared bracket: expand it, spot $sin^{2} x + cos^{2} x$ and replace it by 1, then factorise.

Worked example

1
Solve $(sin x + 2 cos x)^{2} = 1$ for $0^{\circ} \leq x < 36 0^{\circ}$ .
2
$sin^{2} x + 4 sin x cos x + 4 cos^{2} x = 1$ — Expand: first squared, twice the product, last squared.
3
$1 + 4 sin x cos x + 3 cos^{2} x = 1$ — Split $4 cos^{2} x$ into $cos^{2} x + 3 cos^{2} x$ , and use $sin^{2} x + cos^{2} x = 1$ .
4
$cos x (4 sin x + 3 cos x) = 0$ — Take 1 from both sides and factor out $cos x$ .
5
$cos x = 0 \Rightarrow x = 9 0^{\circ}, 27 0^{\circ}$ — Where the cosine wave crosses the axis.
6
$4 sin x = - 3 cos x \Rightarrow tan x = - \frac{3}{4}$ — Divide by $4 cos x$ . Safe now, because the $cos x = 0$ answers are already found.
7
$x = 143. 1^{\circ}, 323. 1^{\circ}$ — Calculator gives −36.87°. Add 180° and 360°.

Common mistake

Dividing by $cos x$ at the factor stage. It throws away $cos x = 0$ , and with it two answers.

Common mistake

Square rooting $cos^{2} A = 0.4$ to get only $cos A = 0.6325$ . Half the answers are lost. Write ±.

Key idea

Squared ratio: root it, keep ±. Squared bracket: expand it, use $sin^{2} x + cos^{2} x = 1$ , factorise.

Sine models

Temperature over a day, the height of a tide, a seat on a big wheel: each goes up and down in a regular wave. Pure 2 questions model them with a sine curve, and ask you to read real facts from it.

Formula

H = A + B sin (k t + c)^{\circ}

$A$ — the middle level the wave goes up and down from.
$B$ — how far it goes above and below the middle. Maximum $A + B$ , minimum $A - B$ .
$k$ — how fast it repeats: one full wave takes $\frac{360}{k}$ units of time.
$c$ — where the wave starts when $t = 0$ .

Rule

The maximum comes when the sine is 1, so the angle inside is 90°. The minimum comes when the sine is −1, so the angle inside is 270°.

Worked example

1
The temperature in a room is $H = 15 + 4 sin (k t + 30)^{\circ}$ °C, $t$ hours after midnight, $0 \leq t < 24$ . At 4 am it is 18.2 °C, and $0 < k < 30$ . Find $k$ . Then, given $k < 10$ , find the maximum temperature and when it happens.
2
$18.2 = 15 + 4 sin (4 k + 30)$ — 4 am means $t = 4$ . Put in the facts you know.
3
$sin (4 k + 30) = 0.8$ — Take 15, then divide by 4.
4
$4 k + 30 = 53.1 3^{\circ}, 126.8 7^{\circ}$ — Calculator value, then $18 0^{\circ} - 53.1 3^{\circ}$ . The next ones give $k$ above 30.
5
$k = 5.78, 24.22$ — Take 30, then divide by 4. Both fit $0 < k < 30$ .
6
$k = 5.78$ — Given $k < 10$ .
7
$H_{m a x} = 15 + 4 = 19^{\circ} C$ — The biggest sine can be is 1.
8
$5.78 t + 30 = 90 \Rightarrow t = 10.38$ — Set the inside angle to 90°.
9
$0.38 \times 60 = 23 min \Rightarrow 10 : 23$ — Turn the decimal hours into minutes.

Common mistake

Setting $k t = 90$ and forgetting the $+ 30$ . The whole inside angle must be 90°.

Common mistake

Leaving the calculator in radians. The model says degrees, so every angle here is in degrees.

Key idea

Lord Kelvin built a brass machine in 1872 that added many sine waves together to predict tides. Ports used machines like it for nearly a hundred years.

Key idea

Middle $A$ , size $B$ . Maximum $A + B$ when the inside is 90°; minimum $A - B$ when the inside is 270°.

Which method? Read the question

Look at the equation before you start. Its shape tells you the first move.

You see	You do
$tan x$ with $sin x$ or $cos x$	Split tan, multiply by $cos x$ , swap $sin^{2} x$ ; solve the quadratic in $cos x$
A root above 1 or below −1	Reject it and say $- 1 \leq cos x \leq 1$
$cos^{2} (\dots) = k$ or $tan^{2} (\dots) = k$	Square root, keep ±, list every solution
$(sin x + a cos x)^{2} = 1$	Expand, use $sin^{2} x + cos^{2} x = 1$ , factorise
$H = A + B sin (k t + c)$	Put in the facts to find $k$ ; max $A + B$ when the inside is 90°

Rule

"Show all stages of your working" means a calculator-only answer scores 0. Write the quadratic, its roots, and then the angles.

Key idea

Each harder type is a route back to one ratio of one angle. From there, solve it the way you already know.

Practise harder trig equations now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on harder trig equations

Each with a full worked solution.

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.