Linear Trigonometric Equations (Edexcel IGCSE Further Pure Maths): Revision Note

Linear Trigonometric Equations

How do I solve trigonometric equations?

• Trigonometric equations can have an infinite number of solutions

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

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

  • For an equation in $\mathbf{tan}\mathit{x}$

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

• 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 $\mathbf{sin}\mathit{x}\mathbf{=}\mathit{k}$

  • The primary value is $x_1={\sin }^{-1}k$

  • By symmetry, a secondary value is $x_2=180°-{\sin }^{-1}k$

    • 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

    • $x_1\pm 360n°$

    • $x_2\pm 360n°$

    • where  $n=1,2,3,...$  as appropriate

• For the equation $\mathbf{cos}\mathit{x}\mathbf{=}\mathit{k}$

  • The primary value is $x_1={\cos }^{-1}k$ 

  • By symmetry, a secondary value is $x_2=-{\cos }^{-1}k$

    • 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

    • $x_1\pm 360n°$

    • $x_2\pm 360n°$

    • where  $n=1,2,3,...$  as appropriate

• For the equation $\mathbf{tan}\mathit{x}\mathbf{=}\mathit{k}$ 

  • The primary value is $x_1={\tan }^{-1}k$
    

    • $x_1$ might not actually be in the given interval!

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

    • $x\pm 180n°$

    • 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, $\sin x\tan x-3\sin x+\tan x=3$

  • Subtract 3 from both sides

    • $\sin x\tan x-3\sin x+\tan x-3=0$

  • Factorise

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

  • This gives you two basic equations to solve

    • $\sin x=-1$

    • $\tan x=3$

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

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