Linear Trigonometric Equations (Edexcel IGCSE Further Maths)

Spec point — 10H

Linear Trigonometric Equations

How do I solve trigonometric equations?

• Trigonometric equations can have an infinite number of solutions

  • For an equation in $\mathrm{sin}x$ or $\mathrm{cos}x$

    • you can add or subtract 360° (or 2π radians) to each solution to find more solutions

  • For an equation in $\mathrm{tan}x$

    • 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 $\mathrm{sin}x=k$, $\mathrm{cos}x=k$ or $\mathrm{tan}x=k$

• 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 $\mathrm{sin}x=k$

  • The primary value is $x_{1}=\mathrm{sin}^{-1}k$

  • By symmetry, a secondary value is $x_{2}=180^{\circ}-\mathrm{sin}^{-1}k$

    • 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

    • $x_{1}\pm360n^{\circ}$
    • $x_{2}\pm360n^{\circ}$
    • where $n=1,2,3,...$ as appropriate

• For the equation $\mathrm{cos}x=k$

  • The primary value is $x_{1}=\mathrm{cos}^{-1}k$

  • By symmetry, a secondary value is $x_{2}=-\mathrm{cos}^{-1}k$

    • 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

    • $x_{1}\pm360n^{\circ}$
    • $x_{2}\pm360n^{\circ}$
    • where $n=1,2,3,...$ as appropriate

• For the equation $\mathrm{tan}x=k$

  • The primary value is $x_{1}=\mathrm{tan}^{-1}k$

    • $x_{1}$ might not actually be in the given interval!

  • Then all values within the given interval can be found using

    • $x\pm180n^{\circ}$
    • 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, $\mathrm{sin}x\mathrm{tan}x-3\mathrm{sin}x+\mathrm{tan}x=3$

  • Subtract 3 from both sides

    • $\mathrm{sin}x\mathrm{tan}x-3\mathrm{sin}x+\mathrm{tan}x-3=0$

  • Factorise

    • $( \sin x + 1 ) ( \tan x - 3 ) = 0$

  • This gives you two basic equations to solve

    • $\mathrm{sin}x=-1$
    • $\mathrm{tan}x=3$

• Trigonometric identities and/or addition formulae may also be needed

Exam Hint

• 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 $2\mathrm{cos}x=-1$, finding all solutions in the interval $-2π\leq x\leq2π$.

First isolate $\mathrm{cos}x$

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

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!

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

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

$x_{2}=-x_{1}=-\frac{2π}{3}$

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

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

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

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

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

Linear Trigonometric Equations (ax + b)

Spec point — 10H

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 Hint

• If you use substitution and transform the interval

  • remember to convert answers back at the end!

Worked Example Solve the equation $2\mathrm{cos}(2x-30^{\circ})=-1$, finding all solutions in the interval $-360^{\circ}\leq x\leq360^{\circ}.$

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

$\mathrm{Let}u=2x-30$

Rewrite the equation in terms of $u$ and rearrange

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

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

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

$2 ( 360 ) - 30 = 690$

$-750^{\circ}\leq u\leq690^{\circ}$

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

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

Use symmetry of cos function to find the secondary value

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

Sketch the graph of $\mathrm{cos}u$ over the transformed interval

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

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

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

Invert the substitution

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

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

$x=-285^{\circ},-225^{\circ},-105^{\circ},-45^{\circ},75^{\circ},135^{\circ},255^{\circ},315^{\circ}$