Why does a trigonometric equation have infinitely many solutions?

Because the functions are periodic , so the same value recurs at regular intervals for ever. A question therefore always supplies an interval , and you must find every solution inside it and none outside.

How many trigonometric solutions can one quadratic produce?

Often more than two , since each valid value of the function can give several angles within the interval. Sketching the relevant graph is the quickest way to check that you have found them all.

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Trigonometric Equations (Edexcel IGCSE Further Pure Maths): Flashcards

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Why does a trigonometric equation have infinitely many solutions?

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Why does a trigonometric equation have infinitely many solutions?
Because the functions are periodic, so the same value recurs at regular intervals for ever.
A question therefore always supplies an interval, and you must find every solution inside it and none outside.
Complete the secondary values, in degrees, for these two equations:
$\sin x = k gives x_{2} =______° - x_{1}, and \cos x = k gives x_{2} =______$
The completed values are:
$\sin x = k gives x_{2} = 180 ° - x_{1}, and \cos x = k gives x_{2} = - x_{1}$
For $\tan$ there is no separate secondary value at all: every solution comes from the first one.
Once you have the first two solutions, how do you find the others?
Add or subtract whole multiples of $360 °$ to each of them, or $2 π$ if you are working in radians.
Keep going until you leave the given interval, and use $180 °$ instead for $\tan$ , whose period is shorter.
Solve $2 \cos x = - 1$ for $- 2 π \leq x \leq 2 π$ .
Isolating gives $\cos x = - \frac{1}{2}$ , with a primary value of $\frac{2 π}{3}$ and a secondary value of $- \frac{2 π}{3}$ .
Adding and subtracting $2 π$ brings in $- \frac{4 π}{3}$ and $\frac{4 π}{3}$ , so there are four solutions altogether.
How do you solve an equation such as $\cos (2 x - 30 °) = k$ ?
Substitute $u = 2 x - 30 °$ and solve for $u$ instead, which turns it into a basic equation.
Transform the interval in the same way, so that you are looking for $u$ over the right range.
Why must the interval be transformed as well as the equation?
Because $u$ runs over a different range from $x$ , and usually a far wider one.
With $u = 2 x - 30$ and $- 360 ° \leq x \leq 360 °$ , the new range is $- 750 ° \leq u \leq 690 °$ , which holds many more solutions.
True or False?
After substituting $u = 2 x - 30 °$ , the answers you find are the values of $x$ .
False.
They are values of $u$ , and each one has to be converted back using $x = \frac{u + 30}{2}$ .
Stopping at $u$ answers a different question from the one that was asked.
Define quadratic trigonometric equation.
A quadratic trigonometric equation is one containing a squared trigonometric term such as $\sin^{2} x$ , $\cos^{2} x$ or $\tan^{2} x$ .
It is solved like any other quadratic, treating the trigonometric function itself as the variable.
What do you do when an equation contains both $\sin^{2} x$ and $\cos x$ ?
Use the Pythagorean identity to replace the squared term, so that only one function is left.
Here $\sin^{2} x = 1 - \cos^{2} x$ turns it into a quadratic entirely in $\cos x$ .
How can a single letter make a quadratic trigonometric equation easier?
Write $s$ for $\sin x$ , so that $5 \sin^{2} x + 11 \sin x - 12 = 0$ becomes $5 s^{2} + 11 s - 12 = 0$ .
Factorising or the quadratic formula then works exactly as usual, and you put $\sin x$ back afterwards.
Why must you check each solution of the quadratic before going on?
Because $\sin x = k$ and $\cos x = k$ only have solutions when $- 1 \leq k \leq 1$ .
So a perfectly good root of the quadratic, such as $\sin x = - 3$ , produces no angles at all and is discarded.
True or False?
$\tan x = 5$ has solutions, even though $5$ is greater than $1$ .
True.
The tangent graph is unbounded, so $\tan x = k$ has solutions for every real value of $k$ .
That is the real difference between the three functions, and it is why a $\tan$ root is never rejected for being too large.
Solve $11 \sin x - 7 = 5 \cos^{2} x$ for $0 \leq x \leq 2 π$ .
Replacing $\cos^{2} x$ with $1 - \sin^{2} x$ gives $5 \sin^{2} x + 11 \sin x - 12 = 0$ , which factorises as $(5 \sin x - 4) (\sin x + 3) = 0$ .
Discarding $\sin x = - 3$ leaves $\sin x = \frac{4}{5}$ , so $x = 0.927$ or $2.21$ to three significant figures.
How many trigonometric solutions can one quadratic produce?
Often more than two, since each valid value of the function can give several angles within the interval.
Sketching the relevant graph is the quickest way to check that you have found them all.

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