KBAT: Strategy for Trig Equations

Q: Why should I not divide both sides by \cos x?

Because cos⁡x\cos xcosx can be zero, and dividing by something that may be zero throws away solutions. In 2sin⁡xcos⁡x=cos⁡x2\sin x\cos x=\cos x2sinxcosx=cosx, dividing gives only sin⁡x=12\sin x=\tfrac{1}{2}sinx=21 and loses x=90∘x=90^{\circ}x=90∘ and 270∘270^{\circ}270∘. Move everything to one side and factor out cos⁡x\cos xcosx instead, then set each factor to zero.

Q: How do I know which identity to use?

Read the mismatch. Two different multiples of the angle, 2x2x2x and xxx, or xxx and x2\tfrac{x}{2}2x, call for a double-angle identity. A squared term with a first-power term calls for sin⁡2x+cos⁡2x=1\sin^{2}x+\cos^{2}x=1sin2x+cos2x=1 to reach one ratio. A lone tan⁡x\tan xtanx among sines and cosines can be written as sin⁡xcos⁡x\tfrac{\sin x}{\cos x}cosxsinx.

Q: How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Choosing the identity, rewriting in one angle, factoring correctly and listing all solutions in the range each earn credit, so show every step and finish with the full set of angles.

What makes this a KBAT question

A routine trig equation is already in a form you can solve: ' $\sin x=0.5$ , find $x$ '. A strategy-choice KBAT question gives you an equation that is not ready, a double angle sitting beside a single angle, a squared term, a product equal to zero, and asks you to decide the route.

That is higher-order thinking, Kemahiran Berfikir Aras Tinggi: every identity is familiar, but you must read what makes the equation awkward and pick the tool that removes it. Two different multiples of the angle call for a double-angle identity; a square calls for the Pythagorean identity; a common factor calls for factoring, never dividing.

Nothing tells you which. In Add Math this rewards students who plan a route before calculating, not only students who can turn a handle once the equation is tidy.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it before reading the solution.

Q1[6 marks]

Solve the equation $\sin 2x=\cos x$ for $0^{\circ}\le x\le 360^{\circ}$ .

Show worked solution

Understand. The left side has a double angle, $\sin 2x$ , while the right side has a single angle, $\cos x$ .

They cannot be compared directly, so we need everything in terms of the same angle $x$ . The double-angle identity $\sin 2x=2\sin x\cos x$ does exactly that.

The range is a full turn, so we should expect several solutions.

Plan. Replace $\sin 2x$ with $2\sin x\cos x$ , move everything to one side, and factor out the common $\cos x$ .

Then set each factor to zero and solve within the range. We will not divide both sides by $\cos x$ , because that would discard any solution where $\cos x=0$ .

Execute and check. Rewrite the left side and bring the right side across:

2\sin x\cos x=\cos x \;\Rightarrow\; 2\sin x\cos x-\cos x=0

Factor out the common factor $\cos x$ :

\cos x\,(2\sin x-1)=0

So either $\cos x=0$ or $2\sin x-1=0$ , that is $\sin x=\tfrac{1}{2}$ . Take each factor in turn over $0^{\circ}\le x\le 360^{\circ}$ .

\cos x=0 \;\Rightarrow\; x=90^{\circ},\;270^{\circ}

For $\sin x=\tfrac{1}{2}$ , the basic angle is $30^{\circ}$ ; sine is positive in the first and second quadrants, so

x=30^{\circ} \quad\text{and}\quad x=180^{\circ}-30^{\circ}=150^{\circ}

Collecting every solution in the range: $x=30^{\circ},\;90^{\circ},\;150^{\circ},\;270^{\circ}$ .

Check. Test $x=90^{\circ}$ : left $\sin 180^{\circ}=0$ , right $\cos 90^{\circ}=0$ , equal.

Test $x=30^{\circ}$ : left $\sin 60^{\circ}=\tfrac{\sqrt{3}}{2}$ , right $\cos 30^{\circ}=\tfrac{\sqrt{3}}{2}$ , equal. Notice that if we had divided by $\cos x$ we would have found only $30^{\circ}$ and $150^{\circ}$ and lost $90^{\circ}$ and $270^{\circ}$ , which is why factoring matters.

Finding a sensible first step

When a trig equation looks unfamiliar, do not start solving, start diagnosing. Scan for what stops the equation being a simple 'ratio equals number'.

If two different multiples of the angle appear, such as $\sin 2x$ and $\cos x$ , the first move is a double-angle identity to make every angle the same. If a squared term appears, such as $\sin^{2}x$ beside $\cos x$ , use $\sin^{2}x+\cos^{2}x=1$ to reach one ratio, then treat it as a quadratic.

If, after rearranging, a product equals zero, factor and set each bracket to zero. The single habit that protects marks: never divide both sides by a trig term that could be zero, move it across and factor instead.

Choose the strategy from the mismatch, and the algebra follows on its own.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a trig equation a marker looks for:

The right identity chosen for the mismatch, double angle for $2x$ beside $x$ , Pythagorean identity for a square.
Every term rewritten in a single angle before solving.
Factoring out the common term rather than dividing, so no solution is lost.
The basic angle found from the positive value, then the correct quadrants used.
All solutions listed within the stated range $0^{\circ}\le x\le 360^{\circ}$ , with none outside it.
A check by substituting one or two of the solutions back into the original equation.

How a teacher helps

The hard part of these questions is the first decision, which identity, and whether to factor, so that is what our teachers rehearse. In a one-to-one lesson we build a short diagnosis routine together: spot the mismatch, name the identity that removes it, then factor rather than divide.

We drill the range work too, so that the basic angle and the quadrant signs give every solution, not just the first. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Why should I not divide both sides by

\cos x

Because $\cos x$ can be zero, and dividing by something that may be zero throws away solutions. In $2\sin x\cos x=\cos x$ , dividing gives only $\sin x=\tfrac{1}{2}$ and loses $x=90^{\circ}$ and $270^{\circ}$ .

Move everything to one side and factor out $\cos x$ instead, then set each factor to zero.

How do I know which identity to use?

Read the mismatch. Two different multiples of the angle, $2x$ and $x$ , or $x$ and $\tfrac{x}{2}$ , call for a double-angle identity.

A squared term with a first-power term calls for $\sin^{2}x+\cos^{2}x=1$ to reach one ratio. A lone $\tan x$ among sines and cosines can be written as $\tfrac{\sin x}{\cos x}$ .

How do I get every solution in the range?

Find the basic (reference) angle from the positive value, then use the sign to choose the quadrants: $\sin x=\tfrac{1}{2}$ is positive in the first and second quadrants, giving $30^{\circ}$ and $150^{\circ}$ . Sweep from $0^{\circ}$ to $360^{\circ}$ and list every matching angle, checking none falls outside the range.

How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Choosing the identity, rewriting in one angle, factoring correctly and listing all solutions in the range each earn credit, so show every step and finish with the full set of angles.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

KBAT: Choosing a Strategy for Trig Equations