Quadratic Trigonometric Equations

How do I solve quadratic trigonometric equations?

A quadratic trigonometric equation is one that includes either $\sin^{2} x$ , $\cos^{2} x$ or $\tan^{2} x$
Often the identity $\sin^{2} θ + \cos^{2} θ = 1$ 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 ${2 \cos}^{2} x + 5 \cos x - 3 = 0$ as $2 c^{2} + 5 c - 3 = 0$
A quadratic can give up to two solutions
- You must check whether solutions to the quadratic are valid solutions
  - ${2 \cos}^{2} x + 5 \cos x - 3 = (2 \cos x - 1) (\cos x + 3) = 0$
  - So $\cos x = \frac{1}{2}$ and $\cos x = - 3$ are the solutions of the quadratic
- Remember that solutions for $\sin x = k$ and only exist for
  - So $\cos x = - 3$ may be a correct solution for the quadratic
  - But it does not give a valid solution for the trigonometric equation!
- Solutions for $\tan x = k$ 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 $\cos x = \frac{1}{2}$
- 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

Exam Tip

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 $11 \sin x - 7 = 5 \cos^{2} x$ , finding all solutions in the interval $0 \leq x \leq 2 π$ . Give your answers correct to 3 significant figures.

$\sin^{2} x + \cos^{2} x = 1$ can be rearranged as $\cos^{2} x = 1 - \sin^{2} x$

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

$11 \sin x - 7 = 5 (1 - \sin^{2} x)$

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

$\begin{array}{rcl} 11 \sin x - 7 & = & 5 - 5 \sin^{2} x \\ 5 \sin^{2} x + 11 \sin x - 12 & = & 0 \end{array}$

This can be solved by factorising (it might help you to think of it as $5 s^{2} + 11 s - 12 = 0$ )
You could also solve it by using the quadratic formula
Or your calculator may be able to solve quadratics

$(5 \sin x - 4) (\sin x + 3) = 0 \sin x = \frac{4}{5} or \sin x = - 3$

$\sin x = - 3$ has no solutions for $x$ because sine cannot be less than $- 1$
So we only need to find solutions for $\sin x = \frac{4}{5}$

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!

$x_{1} = \sin^{- 1} (\frac{4}{5}) = 0.927295 . . .$

Use symmetry properties of sine to find the secondary solution

$x_{2} = π - x_{1} = π - \sin^{- 1} (\frac{4}{5}) = 2.214297 . . .$

Both those solutions are the interval $0 \leq x \leq 2 π$ , and there are no other solutions in the interval
(You could sketch the sine function to confirm that)

$x = 0.927, 2.21 (3 s . f .)$

Quadratic Trigonometric Equations (Edexcel IGCSE Further Maths)

Revision Note