Solving Trig Equations (AQA GCSE Further Maths)

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Paul

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Solving Basic Trig Equations

What are basic trig equations?

The types of equations dealt with here are those of the form
- $\sin x = k$
- $\cos x = k$
- $\tan x = k$
- where $k$ is a constant
(Most) scientific calculators will give you an answer to this type of equation
- using inverse sin/cos/tan
  - e.g. $x = \sin^{- 1} (k)$
- but trig equations usually have more than one solution
For inverse sin (sin^-1) and inverse tan (tan^-1)
- (most) calculators will give an answer between -90° and 90°
For inverse cos (cos^-1)
- (most) calculators will give an answer between 0° and 180°
A question may require answers beyond these ranges
- Typically, answers between 0° and 360° are required
- The properties (in particular, symmetry) of the trig graphs allow us to find all answers to trig equations

How are trigonometric equations of the form sin x = k solved?

The solutions to the equation sin x = 0.5 in the range 0° < x < 360° are x = 30° and x = 150°
- If you like, check on a calculator that both sin 30° and sin 150° give 0.5
The first solution comes from your calculator (by taking inverse sin of both sides)
- x = sin^-1(0.5) = 30°
The second solution comes from the symmetry of the graph y = sin x between 0° and 360°
- Sketch the graph
- Draw a vertical line from x = 30° to the curve, then horizontally across to another point on the curve, then vertically back to the x-axis again
- By the symmetry of the curve, the new value of x is 180° - 30° = 150°
In general, if x° is an acute angle that solves sin x = k, then 180° - x° is the obtuse angle that solves the same equation
If/when a calculator gives x as a negative value when using inverse sin
- extend your graph sketch for negative values of x (i.e. to the left of the y-axis)

cie-igcse-3-12-2-solving-trig-equations-1

How are trigonometric equations of the form cos x = k solved?

The solutions to the equation cos x = 0.5 in the range 0° < x < 360° are x = 60° and x = 300°

If you like, check on a calculator that both cos 60° and cos 300° give 0.5

The first solution comes from your calculator (by taking inverse cos of both sides)

x = cos^-1(0.5) = 60°

The second solution comes from the symmetry of the graph y = cos x between 0° and 360°

Sketch the graph
Draw a vertical line from x = 60° to the curve, then horizontally across to another point on the curve, then vertically back to the x-axis again
By the symmetry of the curve, the new value of x is 360° - 60° = 300°

In general, if x° is an angle that solves cos x = k, then 360° - x° is another angle that solves the same equation
(Most) calculators will not give x as a negative value when using inverse cos

cie-igcse-3-12-2-solving-trig-equations-2

How are trigonometric equations of the form tan x = k solved?

The solutions to the equation tan x = 1 in the range 0° < x < 360° are x = 45° and x = 225°

Check on a calculator that both tan 45° and tan 225° give 1

The first solution comes from your calculator (by taking inverse tan of both sides)

x = tan^-1(1) = 45°

The second solution comes from the symmetry of the graph y = tan x between 0° and 360°

Sketch the graph
Draw a vertical line from x = 45° to the curve, then horizontally across to another point on the curve (a different “branch” of tan x), then vertically back to the x-axis again
The new value of x is 45° + 180° = 225° as the next “branch” of tan x is shifted 180° to the right

In general, if x° is an angle that solves tan x = k, then x° + 180° is another angle that solve the same equation
If/when a calculator gives x as a negative value when using inverse tan
- extend your graph sketch for negative values of x (i.e. to the left of the y-axis)

Exam Tip

Use a calculator to check your solutions by substituting them into the original equation
- e.g. 60° is a correct solution of cos x = 0.5 as cos 60° = 0.5 on a calculator
  but 330° is an incorrect solution as cos 330° ≠ 0.5

Worked example

Solve sin x = 0.25 in the range 0° < x < 360°, giving your answers correct to 1 decimal place

Use a calculator to find the first solution (by taking inverse sin of both sides)

x = sin^-1(0.25) = 14.4775… = 14.48° to 2 dp

Sketch the graph of y = sin x and mark on (roughly) where x = 14.48 and y = 0.25 would be
Draw a vertical line up to the curve, then horizontally across to the next point on the curve, then vertically back down to the x-axis

Find this value using the symmetry of the curve (by taking 14.48 away from 180)

180° – 14.48° = 165.52°

Give both answers correct to 1 decimal place

x = 14.5° or x = 165.5°

Solving Quadratic Trig Equations

What is a quadratic trig equation?

A quadratic trig equation will have a 'squared' term involved in it
- this will be either sin²x, cos²x or tan²x
- there may be terms involving sin x, cos x or tan x as well
  - and possibly a constant term (i.e. a number)
Quadratic trig equations are very similar to normal quadratic equations
- but the 'x' is replaced by 'sin x' (or 'cos x' or 'tan x')
Recall from the work on trig identities that sin²x means the same as (sin x)²
- and likewise for cos/tan

How do I solve quadratic trig equations?

When first practising solving quadratic trig equations it may be helpful to replace sin x, cos x or tan x by a letter, y say
- e.g. for the quadratic trig equation $\tan^{2} x = 8 \tan x - 7$ replacing "tan x" with "y" gives $y^{2} = 8 y - 7$
- if you know what "hidden quadratics" are, this is the same process
Solve the quadratic equation as you normally would
- the equation may need rearranging so it is in quadratic form
  - $" a x^{2} + b x + c = 0 "$
  - e.g. $y^{2} - 8 y + 7 = 0$
- solving the equation may involve factorising, completing the square or the quadratic formula
  - - e.g. $(y - 7) (y - 1) = 0$
      $y = 7, y = 1$
The answer(s) to the quadratic equation will not be the values for x (they are the values of y)
- i.e. they are the values of sin x, cos x or tan x
  - y = tan x in our example
Replace y with its trig function to form two basic trig equations to solve to find the values of x
- e.g. $\tan x = 7$ and $\tan x = 1$
- Using inverse tan, the values for x in the range $0 ° \leq x ° \leq 180 °$ are $x = 81.9$ (1 d.p.) and $x = 45$
- Sketching the graph of $y = \tan x$ between 0° and 360° shows the other two solutions
  - $x = 81.9 + 180 = 261.9 °$ and $x = 45 + 180 = 225 °$
In harder problems, one of the basic trig equations may have no possible solutions
- e.g. $\cos x = 2$
  - The graph y = cos x is always between -1 to 1 on the y-axis, so this trig equation has no solutions

Worked example

Solve the equation $4 \sin^{2} x - 1 = 0$ for $0 ° \leq x ° \leq 360 °$ , giving your answers for x as exact values.

Let y = sin x and rewrite the equation in y

$4 y^{2} - 1 = 0$

This equation can be solved by rearranging to $y^{2} = \frac{1}{4}$ but you must remember $\pm \sqrt{}$ )
A neater way to get two solutions is to use the difference of two squares

$(2 y - 1) (2 y + 1) = 0$

Solve the factorised equation (by setting each bracket equal to zero)

$y = \frac{1}{2}$ or $y = - \frac{1}{2}$

Replace the y with sin x

$\sin x = \frac{1}{2}$ or $\sin x = - \frac{1}{2}$

Write out the "first half" of this problem and solve it

$\sin x = \frac{1}{2}$ where $0 ° < x < 360 °$

Use a calculator (or your knowledge of exact trig values) to find inverse sin

$\sin^{- 1} (\frac{1}{2}) = 30$

Use a sketch of $y = \sin x$ to find the other solution (by symmetry

cie-igcse-3-12-2-solving-trig-equations-1 Write out the two solutions so far

x = 30° or x = 150°

Now write out the "second half" of this problem and solve it

$\sin x = - \frac{1}{2}$ where $0 ° < x < 360 °$

From extending the sketch above in the negative x-axis, or using $\sin^{- 1} (- \frac{1}{2})$ , the first negative solution is

x = -30°

But this is not in the range 0° to 360°
Adding 30° to 180° and subtracting 30° from 360° would give the correct values

x = 180° + 30° = 210° or x = 360° - 30° = 330°

Combine these two solutions with the two found previously
It is good practice to write final solutions out in numerical order

x = 30, 150, 210, 330

Solving Trig Equations with Identities

How do I use trig identities to help solve trig equations?

In general, trig equations are easiest to solve when they involve one out of sin x, cos x or tan x only
For some equations involving different trig functions, you can often rewrite them back in terms of one only
- To do this, use one (or both) of the identities
  - $\tan θ \equiv \frac{\sin θ}{\cos θ}$ or $\sin^{2} θ + \cos^{2} θ \equiv 1$
The majority of questions will involve the second identity as this is frequently used with quadratic trig equations
- It will be used in one of its rearranged forms
  - $\sin^{2} θ = 1 - \cos^{2} θ$ or $\cos^{2} θ = 1 - \sin^{2} θ$
  - which will replace $\sin^{2} θ$ terms (with cos² ...) and $\cos^{2} θ$ terms (with sin² ...) respectively
  - choose whichever suits the question
  - e.g. The equation $\cos^{2} x - 3 \sin x + 1 = 0$ can be rewritten as $(1 - \sin^{2} x) - 3 \sin x + 1 = 0$
    which involves sin x terms only and so is easier to solve (after some rearranging)
The $\tan θ$ identity is used less frequently and would normally appear in reverse
- e.g. The equations $\frac{\sin x}{\cos x} = \frac{3}{4}$ or $4 \sin x = 3 \cos x$ or $3 \cos x - 4 \sin x = 0$ etc can be rewritten as $\tan x = \frac{3}{4}$
- this can be solved using a calculator and the graph of $y = \tan x$

Exam Tip

Write out both identities for any trig equation questions in case they help
Sketch the appropriate graph
- y = sin x, y = cos x or y = tan x
- use the graph to ensure you find ALL solutions within the given interval
- this is particularly important when a calculator gives an initial negative answer

Worked example

Quadratic Trigonometric Equations Example Solution, A Level & AS Level Pure Maths Revision Notes

Worked example

Solve the equation $3 \cos x = 9 \sin x - 6 \cos x$ for $0 ° \leq x ° \leq 360 °$ .

Add 6cos x to both sides to bring the cos x terms together

$9 \cos x = 9 \sin x$

Divide both sides by 9

$\cos x = \sin x$

By considering the identity $\tan x = \frac{\sin x}{\cos x}$ , divide both sides by cos x

$\frac{\cos x}{\cos x} = \frac{\sin x}{\cos x}$

Cancel the left-hand side (this does not cancel to give zero)

$1 = \frac{\sin x}{\cos x}$

Use the identity $\tan x = \frac{\sin x}{\cos x}$ to rewrite the right-hand side as tan x

$1 = \tan x$

The problem is now to solve the following

$\tan x = 1$ where $0 ° < x < 360 °$

Use $\tan^{- 1} (1)$ , or your knowledge of exact trig values, to find the first x solution

x = 45

Sketch the graph of y = tan x to find the other solution

Write the two solutions in numerical order

x = 45 or x = 225

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Solving Trig Equations (AQA GCSE Further Maths)

Revision Note

What are basic trig equations?

How are trigonometric equations of the form sin x = k solved?

How are trigonometric equations of the form cos x = k solved?

How are trigonometric equations of the form tan x = k solved?

What is a quadratic trig equation?

How do I solve quadratic trig equations?

How do I use trig identities to help solve trig equations?

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Author: Paul