Category Archives: Solving Equations

July 26, 2025 · 8:45 am

General Solutions to Trigonometric Equations

Solve $sinx=\frac{1}{2}$ for $0\le x \le 2\pi$

Sine is positive in the first and second quadrants.

$\begin{equation*}sinx=\frac{1}{2}\end{equation}$

$\begin{equation*}x=\frac{\pi}{6} \text{ and } x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\end{equation}$

But what if we aren’t given a domain for the $x$ values?

Then we need to give general solutions.

For example,

Solve $sinx=\frac{1}{2}$

As you can see from the sketch above, there are infinite solutions.

The sine function has a period of $360^\circ$ , and so if $\frac{\pi}{6}$ is a solution then $\2pi+\frac{\pi}{6}$ is also a solution. This means $\frac{\pi}{6}+2\pi n, n\in\mathbb{Z}$ is a general solution. And we can do the same for the second solution $\frac{5\pi}{6}+2\pi n$ .

In general

$\begin{equation*}sinx=y\end{equation}$

$\begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi-arcsin(y)+2\pi n \end{equation}$

$\begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi(2n+1)-arcsin(y), n \in \mathbb{Z}\end{equation}$

We can turn this into one equation
$\begin{equation*}x=(-1)^n arcsin(y)+n\pi, n \in \mathbb{Z}\end{equation}$

What about cosine?

Solve $cosx=\frac{1}{2}$

Cosine is positive in the first and fourth quadrants (it also has a period of $2\pi$ . The first two (positive) solutions are $\frac{\pi}{3}$ and $2\pi-\frac{\pi}{3}$ .

To generalise, $x=2\pi n+\frac{\pi}{3} \text { and }x=2\pi n -\frac{\pi}{3}$ , which we can make into one equation $x=2\pi n \pm \frac{pi}{3}$

In general

$\begin{equation*}cosx=y\end{equation}$

$\begin{equation*}x=2\pi n \pm arccos(y), n\in\mathbb{Z}\end{equation}$

What about the tangent function? Remember tan has a period of $\pi$ .

Solve $tanx=\sqrt{3}$

First, note that the solutions are all a common distance ( $\pi$ ) apart.

Tan is positive in the first and the third quadrant

$\begin{equation*}tanx=\sqrt{3}\end{equation}$

$\begin{equation*}x=\frac{\pi}{3} \text { and } x=\pi+\frac{\pi}{3}\end{equation}$

Because all of the solutions are $\pi$ radians apart, the general solution is $x=\frac{\pi}{3} \pm \pi$

In general

$\begin{equation*}tanx=y\end{equation}$

$\begin{equation*}x=arctan(y) + n\pi, n\in \mathbb{Z}\end{equation}$

Examples

Solve for all values of $x$ , $tan^2(x)+tan(x)-6=0$

$\begin{equation*}tan^2(x)+tan(x)-6=0\end{equation}$

This is a quadratic equation – we need two numbers that add to $1$ and multiple to $-6$ , $+3 \text { and } -2$

$\begin{equation*}(tan(x)+3)(tan(x)-2))=0\end{equation}$

$\begin{equation*}tan(x)=-3 \text { or } tan(x)=2\end{equation}$

$\begin{equation*}x=arctan(-3)+n\pi \text { or } x=arctan(2)+n\pi, n\in\mathbb{Z}\end{equation}$

Solve $2cos(2x+\frac{\pi}{18})=\sqrt{3}$

$\begin{equation*}2cos(2x+\frac{\pi}{18})=\sqrt{3}\end{equation}$

$\begin{equation*}cos(2x+\frac{\pi}{18})=\frac{\sqrt{3}}{2}\end{equation}$

$\begin{equation*}2x+\frac{\pi}{18}=2n\pi \pm \frac{\pi}{6}\end{equation}$

$\begin{equation*}2x=2n\pi \pm \frac{\pi}{6}-\frac{\pi}{18}\end{equation}$

$\begin{equation*}2x=2n\pi \pm \frac{\pi}{9}\end{equation}$

$\begin{equation*}x=n\pi \pm \frac{\pi}{18}\end{equation}$

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Filed under Algebra, Quadratic, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

Tagged as general solution to trig equations, solving trig equations

July 8, 2025 · 6:17 am

Age Question (Year 8 equation solving)

Eight years ago my father was three times as old as I shall be in five years time. When I was born he was 41 years old. How old am I now?

I always find these age questions a bit weird – a bit of a riddle, and contrived (just so we can solve some equations)

Let $x$ be my age now, and $y$ be my fathers age now.

(1) $\begin{equation*}y=x+41\end{equation*}$

Because my father was 41 when I was born.

(2) $\begin{equation*}y-8=3(x+5)\end{equation*}$

$y-8$ for 8 years ago, and $3(x+5)$ for three times my age in 5 years.

Solve the equations simultaneously. Substitute $y=x+41$ into equation $2$

$\begin{equation*}x+41-8=3(x+5)\end{equation}$

$\begin{equation*}x-33=3x+15\end{equation}$

$\begin{equation*}18=2x\end{equation}$

$\begin{equation*}x=9\end{equation}$

Hence my current age is $9$ .

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Filed under Algebra, Simultaneous Equations, Solving Equations, Year 8 Mathematics

Tagged as age question, simultaneous equations, Solving Equations

July 1, 2025 · 10:35 am

Synthetic Division (for factorising and/or solving polynomials)

I use synthetic division to factorise polynomials with a degree greater than 2. For example, $f(x)=x^3+2x^2-5x-6$

It works best with monic polynomials but can be adapted to non-monic ones (see example below).

The only problem is that you need to find a root to start.

Try the factors of $-6$ i.e. $(-1, 1, 2, -2, 3, -3, 6, -6)$

$f(-1)=(-1)^3+2(-1)^2-5(-1)-6=0$

Hence, $x=-1$ is a root and $(x+1)$ is a factor of the polynomial.

Set up as follows

Bring the first number down

Multiply by the root and place under the second co-efficient

Add down

Repeat the process

The numbers at the bottom $(1, 1, -6)$ are the coefficients of the polynomial factor.

We now know $x^3+2x^2-5x-6=(x+1)(x^2+x-6)$ .

We can factorise the quadratic in the usual way.

$x^2+x-6=(x+3)(x-2)$

Hence $x^3+2x^2-5x-6=(x+1)(x+3)(x-2)$ .

Let’s try a non-monic example

Factorise $6x^4+39x^3+91x^2+89x+30$

I know $-2$ is a root. Otherwise I would try the factors of 30.

Use synthetic division

Because this was non-monic we need to divide our new co-efficients $(6, 27, 37, 15)$ by 6 (the co-efficient of the $x^4$ term)

$x^3+\frac{9}{2}x^2+\frac{37}{6}+\frac{5}{2}$

We now need to go again. I know that $\frac{-3}{2}$ is a root and $(2x+3)$ is a factor.

Our quadratic factor is $x^2+3x+5/3$ , which is $3x^2+9x+5$ .

The quadratic factor doesn’t have integer factors so,

$6x^4+39x^3+91x^2+89x+30=(x+2)(2x+3)(3x^2+9x+5)$

I think this is much quicker than polynomial long division.

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Filed under Algebra, Factorising, Factorising, Fractions, Polynomials, Quadratic, Simplifying fractions, Solving Equations

Tagged as non monic synthetic division, Synthetic division

June 26, 2025 · 7:22 am

Trigonometry Question

I don’t know where I found this question, but it does require algebra and problem solving (as well as right trig and Pythagoras)

From a point $A$ , a lighthouse is on a bearing of $026^\circ$ T and the top of the light house is at an angle of elevation of $20.25^\circ$ . From a point $B$ , the lighthouse is on a bearing of $296^\cric$ T and the top of the lighthouse is at angle of elevation of $10.2^\circ$ . If $A$ and $B$ are 500 metres apart, find the height of the lighthouse.

Let’s draw a diagram.

Let the height of the lighthouse be $h$

We can find the angle between $A$ , the lighthouse, and $B$ by using the base triangle

The red line from $L$ is parallel to the two north lines. Hence $\theta=26^\circ+64^\circ=90^\circ$ (Alternate angles in parallel lines are congruent)

It’s a right triangle so we know

(1) $\begin{equation*}500^2=(AL)^2+(BL)^2\end{equation*}$

We are going to use the other two triangles to find $AL$ and $BL$

Substitute $AL$ and $BL$ into equation $1$

$\begin{equation*}500^2=(\frac{h}{tan(20.25)})^2+(\frac{h}{tan(10.2)})^2\end{equation}$

Solve for $h$ .

$\begin{equation*}500^2=7.35h^2+30.89h^2=38.24h^2\end{equation}$

$\begin{equation*}h^2=6538.3\end{equation}$

$\begin{equation*}h=80.9m\end{equation}$

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Filed under Algebra, Bearings, Pythagoras, Right Trigonometry, Solving Equations, Trigonometry

Tagged as hard trig, trigonometry question

May 15, 2025 · 10:01 am

Perfect Squares

Find all of the positive integers that make the following expression a perfect square.

(1) $\begin{equation*}(x-10)(x+14)\end{equation*}$

Let

$\begin{equation*}(x-10)(x+14)=n^2\end{equation}$

where $n$ is an integer.

Expand and simplify

$\begin{equation*}x^2+4x-140=n^2\end{equation}$

$\begin{equation*}x^2-4x-n^2=140\end{equation}$

Complete the square

$\begin{equation*}(x+2)^2-4-n^2=140\end{equation}$

$\begin{equation*}(x+2)^2-n^2=144\end{equation}$

Factorise (using difference of perfect squares)

$\begin{equation*}(x+2-n)(x+2+n)=144\end{equation}$

Find all of the factors of $144$

$(1,144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18), (9, 16), (12, 12)$

First pair,

$\begin{equation*}x+2-n=1 \tag {1} \end{equation}$

$\begin{equation*}x+2+n=144 \tag {2} \end{equation}$

$2x=141$

$x$ must be an integer.

I then used a spreadsheet

Hence the integers that make $(x-10)(x+14)$ are perfect square are, $10, 11, 13, 18,$ and $35$ .

Let’s try another one,

$(x-6)(x+14)$

(2) $\begin{equation*}(x-6)(x+14)=n^2\end{equation*}$

$\begin{equation*}(x^2+8x-84=n^2\end{equation}$

$\begin{equation*}(x+4)^2-n^2=100\end{equation}$

$\begin{equation*}(x+4-n)(x+4+n)=100\end{equation}$

Factors of 100,

$(1, 100), (2, 50), (4, 25), (5, 20), (10, 10)$

So the possible integers are $6$ and $22$ .

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Filed under Algebra, Arithmetic, Divisibility, Interesting Mathematics, Puzzles, Quadratic, Solving Equations

Tagged as perfect squares

May 3, 2025 · 3:34 am

Geometry Puzzle

Another puzzle from this book

Two ladders are propped up vertically in a narrow passageway between two vertical buildings. The ends of the ladders are 8 metres and 4 metres above the pavement.
Find the height above the ground, $T$ ,

$\Delta ABC \sim \Delta TEC$ and $\Delta DCB \sim \Delta TEB$ (Angle Angle Similarity)

Hence,

(1) $\begin{equation*}\frac{h}{8}=\frac{x}{x+y}\end{equation*}$

(2) $\begin{equation*}\frac{h}{4}=\frac{y}{x+y}\end{equation*}$

From equation $1$ $h=\frac{8x}{x+y}$ and from $2$ $h=\frac{4y}{x+y}$

Hence, $\frac{8x}{x+y}=\frac{4y}{x+y}$

Therefore, $8x=4y$ and $y=2x$

From equation $1$

$\begin{equation*}h=\frac{8x}{3x}=\frac{8}{3}\end{equation}$

Hence $T$ is $2 \frac{2}{3}$ m above the ground.

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Filed under Geometry, Puzzles, Similarity, Simplifying fractions, Solving Equations

Tagged as Puzzle

March 31, 2025 · 7:14 am

Deriving the Quadratic Equation formula

My year 10 students have been learning how to complete the square with the idea of then deriving the quadratic equation formula.

The general equation for a quadratic is $y=ax^2+bx+c$

Completing the square,

$\begin{equation*}ax^2+bx+c\end{equation}$

Factorise out the leading coefficient (i.e. $a$ )

$\begin{equation*}a(x^2+\frac{bx}{a}+\frac{c}{a})\end{equation}$

Half the second term (i.e $\frac{b}{a}$ ) and subtract the square of the second term.

$\begin{equation*}a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})\end{equation}$

Simplify

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a^2})\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}\end{equation}$

Now let’s solve

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}=0\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

$\begin{equation*}x=-\frac{b}{2a}\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

Which is the quadratic equation formula.

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Filed under Algebra, Quadratic, Quadratics, Solving, Solving, Solving Equations

Tagged as completing the square, quadratic equation formula

December 3, 2024 · 6:54 am

Puzzle Page 1

If $\frac{a+b+2c}{a+b-c}=\frac{31}{15}$ , what does $\frac{a+b}{c}$ equal?

$\begin{equation*}15(a+n+2c)=31(a+b-c)\end{equation}$

$\begin{equation*}15a+15b+30c=31a+31b-31c)\end{equation}$

$\begin{equation*}61c=16a+16b)\end{equation}$

$\begin{equation*}\frac{61}{16}=\frac{a+b}{c}\end{equation}$

Two positive numbers are such that their difference, their sum, and their product are in the ratio $2:5:21$ . What is the smaller of the two numbers?

Let $x$ and $y$ be the two numbers. Then

(1) $\begin{equation*}x-y=2k\end{equation*}$

(2) $\begin{equation*}x+y=5k\end{equation*}$

(3) $\begin{equation*}xy=21k\end{equation*}$

Add equation $1$ and $2$ together to eliminate the $y$

$\begin{equation*}2x=7k\end{equation}$

(4) $\begin{equation*}x=\frac{7k}{2}\end{equation*}$

From $2$ $=5k-x$ , substitute for $y$ into equation $3$ .

(5) $\begin{equation*}x(5k-x)=21k\end{equation*}$

Substitute $x=\frac{7k}{2}$ into equation $5$ .

$\begin{equation*}\frac{7k}{2}(5k-\frac{7k}{2})=21k\end{equation}$

$\begin{equation*}\frac{35k^2}{2}-\frac{49k^2}{4}=21k\end{equation}$

$\begin{equation*}\frac{70k^2}{4}-\frac{49k^2}{4}=\frac{84k}{4}\end{equation}$

$\begin{equation*}21k^2-84k=0\end{equation}$

$\begin{equation*}21k(k-4)=0\end{equation}$

Hence, $k=0$ or $k=4$ .

When $k=4$ , $x=\frac{7\times 4}{2}=14$ and $y=5\times 4-14=6$

Therefore the smaller number is $6$ .

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Filed under Algebra, Puzzles, Ratio, Solving Equations

Tagged as problem solving, ratio, Solving Equations

Category Archives: Solving Equations

General Solutions to Trigonometric Equations

Examples

Age Question (Year 8 equation solving)

Synthetic Division (for factorising and/or solving polynomials)

Trigonometry Question

Perfect Squares

Geometry Puzzle

Deriving the Quadratic Equation formula

Puzzle Page 1

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