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SingaporeAdditional MathematicsSyllabus dot point

How do we solve a trigonometric equation and find all the angles in a given range?

Solve simple trigonometric equations within a stated range using the basic angle and the quadrant rule

A focused answer to the N(A)-Level Additional Mathematics outcome on trigonometric equations. Find the basic angle, use the quadrant sign rule to locate all solutions, and list every angle within a stated range.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to solve a trigonometric equation such as sinx=0.5\sin x = 0.5 and find every solution within a stated range (often 00^\circ to 360360^\circ). The method is always the same: find the basic angle from the positive value, decide which quadrants the solutions lie in from the sign, then write down each angle in range. The most common error is giving only the calculator answer and missing the others.

The answer

Step one: find the basic angle

The basic angle (reference angle) is the acute angle you get by taking the inverse trig function of the positive value:

basic angle=sin1v, cos1v, or tan1v\text{basic angle} = \sin^{-1}|v|, \ \cos^{-1}|v|, \ \text{or}\ \tan^{-1}|v|

Always use the positive value here; the sign of the original number is handled separately by the quadrants. For example, for cosx=0.5\cos x = -0.5 the basic angle is cos1(0.5)=60\cos^{-1}(0.5) = 60^\circ.

Step two: choose the quadrants from the sign

The sign of the right-hand side tells you which quadrants the solutions are in, using "All, Sine, Tangent, Cosine":

  • sinx\sin x positive: quadrants 1 and 2; negative: quadrants 3 and 4.
  • cosx\cos x positive: quadrants 1 and 4; negative: quadrants 2 and 3.
  • tanx\tan x positive: quadrants 1 and 3; negative: quadrants 2 and 4.

Step three: build the angles in each quadrant

With basic angle α\alpha, the angle in each quadrant is:

  • Quadrant 1: α\alpha
  • Quadrant 2: 180α180^\circ - \alpha
  • Quadrant 3: 180+α180^\circ + \alpha
  • Quadrant 4: 360α360^\circ - \alpha

Write down the two (or more) that fall in the chosen quadrants and lie within the given range.

Handling a wider range or a multiple angle

If the range is larger (say up to 720720^\circ), keep adding 360360^\circ to each solution while it stays in range. If the equation is in a multiple angle like sin2x\sin 2x, solve for 2x2x over the stretched range first, then divide each answer by 22.

Examples in context

Example 1. After applying an identity. A question may first need an identity to reduce 2sin2x=12\sin^2 x = 1 to sinx=±12\sin x = \pm\tfrac{1}{\sqrt{2}}, which then splits into two standard equations. So the identity work and the equation-solving method combine in fuller questions.

Example 2. Modelling periodic motion. The height of a point on a turning wheel, or the depth of a tide, follows a sine model. Finding the times when the height takes a particular value means solving a trigonometric equation over a time range, which is exactly this skill applied to a real periodic situation.

Try this

Q1. Solve cosx=0.5\cos x = 0.5 for 0x3600^\circ \le x \le 360^\circ. [2 marks]

  • Cue. Basic angle 6060^\circ; cosine positive in quadrants 1 and 4, so x=60x = 60^\circ and x=300x = 300^\circ.

Q2. Solve tanx=1\tan x = 1 for 0x3600^\circ \le x \le 360^\circ. [2 marks]

  • Cue. Basic angle 4545^\circ; tangent positive in quadrants 1 and 3, so x=45x = 45^\circ and x=225x = 225^\circ.

Q3. Solve sinx=0\sin x = 0 for 0x3600^\circ \le x \le 360^\circ. [2 marks]

  • Cue. sinx=0\sin x = 0 at x=0x = 0^\circ, 180180^\circ and 360360^\circ.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksSolve sinx=0.5\sin x = 0.5 for 0x3600^\circ \leq x \leq 360^\circ.
Show worked answer →

Basic angle: sin1(0.5)=30\sin^{-1}(0.5) = 30^\circ.

Sine is positive in the first and second quadrants, so x=30x = 30^\circ and x=18030=150x = 180^\circ - 30^\circ = 150^\circ.

What markers reward: finding the basic angle 3030^\circ, recognising sine is positive in quadrants 1 and 2, and giving both solutions 3030^\circ and 150150^\circ within the range.

Original3 marksSolve cosx=0.5\cos x = -0.5 for 0x3600^\circ \leq x \leq 360^\circ.
Show worked answer →

Basic angle: cos1(0.5)=60\cos^{-1}(0.5) = 60^\circ (use the positive value for the basic angle).

Cosine is negative in the second and third quadrants, so x=18060=120x = 180^\circ - 60^\circ = 120^\circ and x=180+60=240x = 180^\circ + 60^\circ = 240^\circ.

What markers reward: taking the basic angle from the positive value, placing solutions in the quadrants where cosine is negative, and giving 120120^\circ and 240240^\circ.

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