Sign In | Starter Of The Day | Tablesmaster | Fun Maths | Maths Map | Topics | More

International Baccalaureate Mathematics

Geometry and Trigonometry

Syllabus Content

Solving trigonometric equations in a finite interval, both graphically and analytically. Equations leading to quadratic equations in sinx, cosx or tanx

Here are some specific activities, investigations or visual aids we have picked out. Click anywhere in the grey area to access the resource.

Here are some exam-style questions on this statement:

See all these questions

Here are some Advanced Starters on this statement:

Click on a topic below for suggested lesson Starters, resources and activities from Transum.


Furthermore

Official Guidance, clarification and syllabus links:

$$2\sin x = 1, \quad 0 \leq x \leq 2\pi$$

Examples:
$$2\sin 2x = 3\cos x, \quad 0^{\circ} \leq x \leq 180^{\circ}$$
$$2\tan(3(x - 4)) = 1, \quad - \pi \leq x \leq 3\pi$$

$$2\sin^2 x + 5\cos x + 1 = 0 \text{ for } 0 \leq x \leq 4\pi,$$
$$2\sin x = \cos 2x, \quad - \pi \leq x \leq \pi$$

Not required: The general solution of trigonometric equations.

Solving trigonometric equations within a finite interval involves finding all the angles that satisfy the equation within a given range. This can be approached graphically, by plotting the functions and identifying the points of intersection, or analytically, by manipulating the equation using trigonometric identities and inverse functions. When trigonometric equations reduce to a quadratic form in \(\sin x\), \(\cos x\), or \(\tan x\), we can solve them using algebraic techniques similar to those used for quadratic equations.

Key Formulae:
$$\sin^2 x + \cos^2 x = 1$$
$$1 + \tan^2 x = \sec^2 x$$
$$\sin(2x) = 2\sin x \cos x$$
$$\cos(2x) = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$$

Example:
To solve the equation \(2\sin^2 x - \sin x - 1 = 0\) for \(0 \leq x \leq 2\pi\):
Let \(u = \sin x\). The equation becomes \(2u^2 - u - 1 = 0\).
Solving this quadratic equation, we find \(u = -\frac{1}{2}\) or \(u = 1\).
Returning to \(x\), we have \(\sin x = -\frac{1}{2}\) or \(\sin x = 1\).
This yields the solutions \(x = \frac{7\pi}{6}, \frac{11\pi}{6}\) or \(x = \frac{\pi}{2}\) within the given interval.

This video on Solving Trig Functions and Equations is from Revision Village and is aimed at students taking the IB Maths AA SL/HL course.

How do you teach this topic? Do you have any tips or suggestions for other teachers? It is always useful to receive feedback and helps make these free resources even more useful for Maths teachers anywhere in the world. Click here to enter your comments.


Apple

©1997-2024 WWW.TRANSUM.ORG