Linear Trigonometric Equations

How do I solve trigonometric equations?

Trigonometric equations can have an infinite number of solutions
- For an equation in $\sin x$ or $\cos x$
  - you can add or subtract 360° (or 2π radians) to each solution to find more solutions
- For an equation in $\tan x$
  - you can add or subtract 180° (or π radians) to each solution to find more solutions
When solving a trigonometric equation
- You will be given a interval of values within which the answers must lie
  - You need to find all the answers within that range
- Using the inverse function on your calculator will only give you the primary value
  - This may or may not be in the required interval
- The other values can be found with the help of:
  - your knowledge of trigonometric exact values
  - the unit circle
  - the graphs of trigonometric functions

How are basic trigonometric equations solved?

This means equations in the form $\sin x = k$ , $\cos x = k$ or $\tan x = k$
It can be helpful to sketch the graph of the trigonometric function first
- Use the given interval of values as the domain for your graph
- The intersections of the graph of the function and the line $y = k$ will show you
  - The location of the solutions
  - The number of solutions
- You will be able to use the symmetry properties of the graph to find other values within the given interval

The methods for finding all solutions are:

For the equation $\sin x = k$
- The primary value is $x_{1} = \sin^{- 1} k$
- By symmetry, a secondary value is
  - Either $x_{1}$ or $x_{2}$ might not actually be in the given interval!
- Then all values within the given interval can be found using
  - $x_{1} \pm 360 n °$
  - $x_{2} \pm 360 n °$
  - where $n = 1, 2, 3, . . .$ as appropriate
For the equation $\cos x = k$
- The primary value is $x_{1} = \cos^{- 1} k$
- By symmetry, a secondary value is
  - Either $x_{1}$ or $x_{2}$ might not actually be in the given interval!
- Then all values within the given interval can be found using
  - $x_{1} \pm 360 n °$
  - $x_{2} \pm 360 n °$
  - where $n = 1, 2, 3, . . .$ as appropriate
For the equation $\tan x = k$
- The primary value is
  - $x_{1}$ might not actually be in the given interval!
- Then all values within the given interval can be found using
  - $x \pm 180 n °$
  - where $n = 1, 2, 3, . . .$ as appropriate

How do I handle more complicated equations?

You may need to use algebra to get an equation into one of the basic forms
For example,
- Subtract 3 from both sides
  - $\sin x \tan x - 3 \sin x + \tan x - 3 = 0$
- Factorise
  - $(\sin x + 1) (\tan x - 3) = 0$
- This gives you two basic equations to solve
  - $\sin x = - 1$
  - $\tan x = 3$
Trigonometric identities and/or addition formulae may also be needed

Exam Tip

Remember that your calculator will only give you the primary value
- You need to be able to find all other solutions within the given interval
Sketching the trig graphs (or any other useful diagrams) can be a huge help!

Worked example

Solve the equation $2 \cos x = - 1$ , finding all solutions in the interval $- 2 π \leq x \leq 2 π$ .

First isolate $\cos x$

$\cos x = - \frac{1}{2}$

Use calculator or knowledge of exact trig values to find $x_{1}$
Note that the interval is given in radians, so we must work in radians!

$x_{1} = \cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$

Use symmetry of the cos function to find $x_{2}$

$x_{2} = - x_{1} = - \frac{2 π}{3}$

Now add or subtract (multiples of) $2 π$ radians to find other solutions in the interval

$x_{1} - 2 π = \frac{2 π}{3} - 2 π = - \frac{4 π}{3}$

$x_{2} + 2 π = - \frac{2 π}{3} + 2 π = \frac{4 π}{3}$

Any other additions or subtractions of $2 π$ would take us outside the interval

$x = - \frac{4 π}{3}, - \frac{2 π}{3}, \frac{2 π}{3}, \frac{4 π}{3}$

Linear Trigonometric Equations (ax + b)

How can I solve equations with transformations of trig functions?

This means equations of the form sin(ax+b) = k, cos(ax+b) = k or tan(ax+b) = k
- Trigonometric equations in these forms can be solved in more than one way
The easiest method is to consider the transformation of the angle as a substitution
- Let u = ax + b
Transform the given interval for the solutions in the same way as the angle
- For example if the given interval is 0° ≤ x ≤ 360° the new interval will be
  - (a (0°) + b) ≤ u ≤ (a (360°) + b)
Solve the equation to find the primary value for u
Find all the other solutions in the transformed range for u
Undo the substitution
- i.e. $x = \frac{u - b}{a}$
- Convert all of the u solutions back into corresponding solutions for x
Another method would be to sketch the transformation of the function
- If you use this method then you will not need to use a substitution for the range of values

Exam Tip

If you use substitution and transform the interval
- remember to convert answers back at the end!

Worked example

Solve the equation $2 \cos (2 x - 30 °) = - 1$ , finding all solutions in the interval $- 360 ° \leq x \leq 360 ° .$

We'll use the substitution $u = 2 x - 30$

$Let u = 2 x - 30$

Rewrite the equation in terms of $u$ and rearrange

$\begin{array}{rcl} 2 \cos u & = & - 1 \\ \cos u & = & - \frac{1}{2} \end{array}$

Now we need to transform the interval as well
Substitute the interval limits into $u = 2 x - 30$

$2 (- 360) - 30 = - 750$

$2 (360) - 30 = 690$

$- 750 ° \leq u \leq 690 °$

Use calculator or knowledge of exact trig values to find the primary value

$u_{1} = \cos^{- 1} (- \frac{1}{2}) = 120$

Use symmetry of cos function to find the secondary value

$u_{2} = - u_{1} = - 120$

Sketch the graph of $\cos u$ over the transformed interval

cxVOKRsE_igcse-fpure-trig-eqns-we

This shows that there are 8 places where $\cos u = - \frac{1}{2}$

Find these by adding or subtracting (multiples of) $360 °$ to $u_{1}$ and $u_{2}$

$u = - 600, - 480, - 240, - 120, 120, 240, 480, 600$

Invert the substitution

$\begin{array}{rcl} u & = & 2 x - 30 \\ 2 x & = & u + 30 \\ x & = & \frac{u + 30}{2} \end{array}$

Substitute the $u$ values into $x = \frac{u + 30}{2}$ to find the corresponding $x$ values

$x = - 285 °, - 225 °, - 105 °, - 45 °, 75 °, 135 °, 255 °, 315 °$

Linear Trigonometric Equations (Edexcel IGCSE Further Maths)

Revision Note

Linear Trigonometric Equations

How do I solve trigonometric equations?

How are basic trigonometric equations solved?

How do I handle more complicated equations?

Exam Tip

Worked example

Linear Trigonometric Equations (ax + b)

How can I solve equations with transformations of trig functions?

Exam Tip

Worked example

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Author: Amber