What are the main ways of solving a trigonometric equation?

There are four: sketching the relevant graph using a trigonometric identity using the CAST diagram factorising a quadratic Which of them is quickest depends on the form the equation is in.

What should you check before writing down your final solutions to a trigonometric equation?

Check that you have found all of them, and that every one lies inside the interval you were given. If you transformed the interval in order to solve, check that the solutions have been transformed back.

ASMathsEdexcelPureFlashcardsTrigonometryTrigonometric Equations

Trigonometric Equations (Edexcel AS Maths: Pure): Flashcards

Exam code: 8MA0

1/19

0Still learning

Know0

FrontLinear Trigonometric Equations
Define the CAST diagram.

Was this flashcard helpful?

Cards in this collection (19)

Define the CAST diagram.
The CAST diagram is a set of four quadrants labelled A, S, T and C, showing which trigonometric functions are positive in each one.
Going anticlockwise from $0 °$ they are A (all three positive), S (sine only), T (tangent only) and C (cosine only).
You have found the principal value of a trigonometric equation. How do you use the CAST diagram to find the other solutions between $0 °$ and $360 °$ ?
Draw the principal value in as an angle, then repeat that same angle in all four quadrants.
Keep only the two quadrants in which your function has the same sign as the value you are solving for.
Read each remaining solution off as the angle measured anticlockwise from $0 °$ .
True or False?
In the CAST diagram, the angle you draw into each quadrant is measured from the nearest vertical axis.
False.
It is always measured from the horizontal axis, in every quadrant.
Measuring from the vertical instead uses $90 °$ minus your angle as the reference, which sends every solution to the wrong place.
The CAST diagram has given you every solution between $0 °$ and $360 °$ , but the interval is wider. How do you find the rest?
Add $360 °$ to each solution you already have, and subtract $360 °$ from it, repeating until the results fall outside the interval.
A whole $360 °$ returns to the same point of the cycle, so the function takes the same value there.
$\sin 2 x = \frac{1}{\sqrt{2}}$ is to be solved for $0 ° \leq x \leq 360 °$ using the substitution $Z = 2 x$ . Complete the interval that $Z$ runs through:
$______\leq Z \leq______$
The completed interval is:
$0 ° \leq Z \leq 720 °$
Both ends are multiplied by $2$ , so $Z$ runs through two full cycles and there are twice as many solutions to find.
An equation in $\cos (θ - 30 °)$ is to be solved for $0 ° \leq θ \leq 360 °$ . What interval do you solve it in, after substituting $Z = θ - 30 °$ ?
The interval is $- 30 ° \leq Z \leq 330 °$ , found by subtracting $30 °$ from both ends.
Apply to the interval exactly what the substitution does to the angle, or you will miss some solutions and keep others that do not belong.
After solving a trigonometric equation for $Z$ , where $Z = 2 x$ , how do you turn your answers into values of $x$ ?
Convert each one by halving it, since $Z = 2 x$ means $x = \frac{Z}{2}$ .
For a substitution such as $Z = θ - 30 °$ you would instead add $30 °$ to each solution.
What makes a trigonometric equation a quadratic one?
It contains the square of a trigonometric function, such as $\sin^{2} θ$ or $\cos^{2} θ$ .
Once every term is written using the same function, it has the form $a x^{2} + b x + c = 0$ and is solved like any other quadratic.
Factorising a quadratic trigonometric equation gives $(2 \cos θ + 1) (\cos θ - 2) = 0$ . Complete the two equations this produces:
$\cos θ =______$
$\cos θ =______$
The two equations are:
$\cos θ = - \frac{1}{2}$
$\cos θ = 2$
Writing $C$ for $\cos θ$ while you factorise makes the quadratic easier to see, provided you put $\cos θ$ back afterwards.
A quadratic trigonometric equation has two roots. Does that mean it has two solutions in the given interval?
No: each root is a separate equation such as $\sin x = \frac{4}{5}$ , and each can have several solutions inside the interval, or none at all.
The number of solutions is decided by the interval, not by the number of roots.
True or False?
$\tan x = k$ has a solution for every value of $k$ , however large.
True.
The tangent function is unbounded, so every horizontal line crosses its graph and no value of $k$ is out of reach.
So a root such as $\tan x = 7$ is perfectly solvable, even though it looks too big.
Factorising a quadratic trigonometric equation gives the roots $\cos θ = - \frac{1}{2}$ and $\cos θ = 2$ . What do you do with the second root?
Discard it, because $\cos θ$ can never be greater than $1$ , so $\cos θ = 2$ has no solutions.
Every solution of the original equation then comes from $\cos θ = - \frac{1}{2}$ alone.
This quadratic trigonometric equation has no constant term. Complete its factorisation:
$\tan^{2} x - 2 \tan x =______(\tan x - 2)$
The completed factorisation is:
$\tan^{2} x - 2 \tan x = \tan x (\tan x - 2)$
Taking the common factor out keeps the equation $\tan x = 0$ , which supplies the solutions $0 °$ , $180 °$ and $360 °$ .
Solving $\sin x = - \frac{1}{4}$ for $0 ° \leq x \leq 360 °$ , your calculator gives $- 14.5 °$ . Where are the actual solutions?
The solutions are $180 ° + 14.5 ° = 194.5 °$ and $360 ° - 14.5 ° = 345.5 °$ .
A negative value from the calculator is still useful: it is not itself in the interval, but it is the angle everything else is measured from.
What are the main ways of solving a trigonometric equation?
There are four:
- sketching the relevant graph
- using a trigonometric identity
- using the CAST diagram
- factorising a quadratic
Which of them is quickest depends on the form the equation is in.
Before you start solving, what is the first thing to check about the angle in a trigonometric equation?
The first thing to check is whether the angle is a function of the unknown, such as $2 x$ or $θ + 30 °$ , rather than the unknown on its own.
If it is, the interval has to be transformed to match before you solve anything.
Before any trigonometry can start, $3 + 5 \cos 2 x = 1$ has to be rearranged. Complete it:
$\cos 2 x =______$
The rearranged equation is:
$\cos 2 x = - \frac{2}{5}$
Subtract $3$ from both sides, then divide by $5$ : the trigonometric function must be on its own before an inverse can be taken.
True or False?
One trigonometric equation can need both an identity and a factorisation before it is ready to solve.
True.
In $6 \cos^{2} x + \sin x - 5 = 0$ an identity turns every term into sine, giving $6 \sin^{2} x - \sin x - 1 = 0$ , and only then can it be factorised.
The methods are steps that chain together, not alternatives you choose between.
What should you check before writing down your final solutions to a trigonometric equation?
Check that you have found all of them, and that every one lies inside the interval you were given.
If you transformed the interval in order to solve, check that the solutions have been transformed back.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Trigonometric FunctionsNext:Exponential & Logarithms