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Quadratic Trigonometric Equations (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Quadratic Trigonometric Equations

How do I solve quadratic trigonometric equations?

  • A quadratic trigonometric equation is one that includes either sin squared space x, cos squared space x or tan to the power of 2 space end exponent x

  • Often the identityspace sin squared space theta plus cos squared space theta equals 1 can be used to help solve the equation

    • This can change an equation with both sine and cosine

    • into an equation with only sine or cosine

  • Solve the quadratic equation using any of the usual methods

    • You may find it easier to rewrite it as an equation with a single letter

      • e.g. writing  2 cos squared space x plus 5 cos space x minus 3 equals 0  as  2 c squared plus 5 c minus 3 equals 0

  • A quadratic can give up to two solutions

    • You must check whether solutions to the quadratic are valid solutions

      • 2 cos squared space x plus 5 cos space x minus 3 equals open parentheses 2 cos x minus 1 close parentheses open parentheses cos x plus 3 close parentheses equals 0

      • So cos x equals 1 half and cos x equals negative 3 are the solutions of the quadratic

    • Remember that solutions for sin x equals k and cos x equals k only exist for negative 1 less or equal than k less or equal than 1

      • So cos x equals negative 3 may be a correct solution for the quadratic

      • But it does not give a valid solution for the trigonometric equation!

    • Solutions for tan x equals k exist for all values of k

  • After you solve the quadratic equation

    • Find all solutions for the resulting trigonometric equation(s) within the given interval

      • For the example above this would mean solving cos x equals 1 half 

    • There will often be more than two trigonometric solutions for one quadratic equation

    • Sketching a graph can help check how many solutions there should be in the given interval

Examiner Tips and Tricks

  • Sketch the trig graphs on your exam paper

    • Then you can refer back to them as many times as you need to

  • Make sure you have found all of the solutions in the given interval

    • And that you don't give solutions outside the interval 

    • For example if you get a negative solution but the interval is entirely positive

Worked Example

Solve the equation 11 sin space x – 7 equals 5 cos squared space x, finding all solutions in the interval  0 less or equal than x less or equal than 2 pi.  Give your answers correct to 3 significant figures.

sin squared x plus cos squared x equals 1  can be rearranged as  cos squared x equals 1 minus sin squared x

Substitute this to get the equation entirely in terms of sin x

11 sin x minus 7 equals 5 open parentheses 1 minus sin squared x close parentheses

Expand the brackets and rearrange to get a quadratic equal to zero

table row cell 11 sin x minus 7 end cell equals cell 5 minus 5 sin squared x end cell row cell 5 sin squared x plus 11 sin x minus 12 end cell equals 0 end table

This can be solved by factorising (it might help you to think of it as  5 s squared plus 11 s minus 12 equals 0)
You could also solve it by using the quadratic formula
Or your calculator may be able to solve quadratics

open parentheses 5 sin x minus 4 close parentheses open parentheses sin x plus 3 close parentheses equals 0 sin x equals 4 over 5 space space or space space sin x equals negative 3

sin x equals negative 3  has no solutions for x because sine cannot be less than negative 1
So we only need to find solutions for sin x equals 4 over 5

Start by finding the primary solution
The interval is given in radians, so we have to make sure the calculator is set up for radians!

x subscript 1 equals sin to the power of negative 1 end exponent open parentheses 4 over 5 close parentheses equals 0.927295...

Use symmetry properties of sine to find the secondary solution

x subscript 2 equals pi minus x subscript 1 equals pi minus sin to the power of negative 1 end exponent open parentheses 4 over 5 close parentheses equals 2.214297...


Both those solutions are the interval 0 less or equal than x less or equal than 2 pi, and there are no other solutions in the interval
(You could sketch the sine function to confirm that)


bold italic x bold equals bold 0 bold. bold 927 bold comma bold space bold 2 bold. bold 21 bold space bold space stretchy left parenthesis 3 space s. f. stretchy right parenthesis

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.