Trigonometric Functions

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

General solutions to trigonometric equations

Lesson

A *general solution* to a trigonometric equation is an expression or set of expressions that represents all possible solutions.

Because of the periodic nature of trigonometric functions, a trigonometric equation will usually have many solutions and these will occur at regular intervals. If we can find one or more basic solutions, the others can be specified by adding multiples of some fixed amount to the known solutions.

For example, the equation $\sin\theta=\frac{1}{\sqrt{3}}$`s``i``n``θ`=1√3 has the solution $\theta_1=\arcsin\frac{1}{\sqrt{3}}$`θ`1=`a``r``c``s``i``n`1√3 in the first quadrant - between $0$0 and $\frac{\pi}{2}$π2. It must also have the second quadrant solution $\theta_2=\pi-\arcsin\frac{1}{\sqrt{3}}$`θ`2=π−`a``r``c``s``i``n`1√3.

We know that the sine function is periodic with period $2\pi$2π. Thus, any multiple of $2\pi$2π added to the first or second quadrant solution will also be a solution. This means further solutions can be expressed as

$\theta_1+2n\pi$`θ`1+2`n`π and

$\theta_2+2n\pi$`θ`2+2`n`π which is the same as $\pi-\theta_1+2n\pi=(2n+1)\pi-\theta_1$π−`θ`1+2`n`π=(2`n`+1)π−`θ`1, where $n$`n` is any positive or negative integer.

Guidance about finding initial solutions to trigonometric equations can be found at linked chapters A and B.

Find the general solution of $\sin\theta=\frac{1}{2}$`s``i``n``θ`=12 in radians.

You may use $n$`n` to represent an integer.

We are being asked to find all the values for $\theta$`θ`, that make $\sin\theta=\frac{1}{2}$`s``i``n``θ`=12 true.

Let's just remind ourselves of the graphical solution.

Now - let's see how to do this analytically (which just means by hand!)

Firstly we should recognise that $\sin\frac{\pi}{6}=\frac{1}{2}$`s``i``n`π6=12.

So we should use $\frac{\pi}{6}$π6 as a reference angle to find all angles in the interval $\left[0,2\pi\right)$[0,2π) whose sine is equal to $\frac{1}{2}$12.

In the interval $\left[0,2\pi\right)$[0,2π), $\theta$`θ` can be both $\frac{\pi}{6}$π6 and $\frac{5\pi}{6}$5π6.

Since $\sin\theta$`s``i``n``θ` has a period of $2\pi$2π, then the general solution will be

$\theta=\alpha+2n\pi$`θ`=`α`+2`n`π or $\theta=\beta+2n\pi$`θ`=`β`+2`n`π

where $\alpha$`α` and $\beta$`β` are all the solutions in the interval $\left[0,2\pi\right)$[0,2π)

So $\alpha$`α` is $\frac{\pi}{6}$π6 and $\beta$`β` is $\frac{5\pi}{6}$5π6

Which means our general solutions are

$\theta_1=\frac{\pi}{6}+2n\pi$`θ`1=π6+2`n`π and $\theta_2=\frac{5\pi}{6}+2n\pi$`θ`2=5π6+2`n`π

Let's just check the graph again, how these general solutions map.

Suppose that in the course of an investigation, we need to solve the following trigonometric equation:

$3\sin2x=4\cos^2x$3`s``i``n`2`x`=4`c``o``s`2`x`

We might begin by writing $\sin2x$`s``i``n`2`x` as $2\sin x\cos x$2`s``i``n``x``c``o``s``x`. Then, the equation is equivalent to $6\sin x\cos x=4\cos^2x$6`s``i``n``x``c``o``s``x`=4`c``o``s`2`x`.

On moving all the terms to the left and factorising, we have, $\cos x\left(6\sin x-4\cos x\right)=0$`c``o``s``x`(6`s``i``n``x`−4`c``o``s``x`)=0.

This is true when $\cos x=0$`c``o``s``x`=0 and when $\left(6\sin x-4\cos x\right)=0$(6`s``i``n``x`−4`c``o``s``x`)=0.

From the factor $\cos x=0$`c``o``s``x`=0, we have $x=\frac{\pi}{2}$`x`=π2 or $x=-\frac{\pi}{2}$`x`=−π2, and when we add $2n\pi$2`n`π to each of these solutions because of the periodicity of the cosine function, we get the general solutions $x=2n\pi+\frac{\pi}{2}$`x`=2`n`π+π2 and $x=2n\pi-\frac{\pi}{2}$`x`=2`n`π−π2 In this case, the difference between successive solutions is just $\pi$π. That is, $2n\pi-\frac{\pi}{2}+\pi=2n\pi+\frac{\pi}{2}$2`n`π−π2+π=2`n`π+π2 and also, $2n\pi+\frac{\pi}{2}+\pi=2(n+1)\pi-\frac{\pi}{2}$2`n`π+π2+π=2(`n`+1)π−π2. So, we can write the solutions more simply as

$x=\frac{\pi}{2}+n\pi$`x`=π2+`n`π.

From the other factor, $6\sin x-4\cos x=0$6`s``i``n``x`−4`c``o``s``x`=0, we can write a simpler equation $6\frac{\sin x}{\cos x}=4$6`s``i``n``x``c``o``s``x`=4. Then, simplifying further, $\tan x=\frac{2}{3}$`t``a``n``x`=23.

Therefore, $x=\arctan\frac{2}{3}$`x`=`a``r``c``t``a``n`23 in the first quadrant. The tangent function has period $\pi$π. So, the set of solutions arising from this one is given by

$x=\arctan\frac{2}{3}+n\pi$`x`=`a``r``c``t``a``n`23+`n`π, for integers $n$`n`.

We could, for practical purposes, give an approximate evaluation of the $\arctan$`a``r``c``t``a``n` expression. Thus, $x\approx0.5880+3.1416\ n$`x`≈0.5880+3.1416 `n`. Again for practicality, we might wish to convert this to degrees. Thus, $x\approx0.588\times\frac{360^\circ}{2\pi}+180^\circ\ n\ \approx33.6899^\circ+180^\circ\ n$`x`≈0.588×360°2π+180° `n` ≈33.6899°+180° `n`.

The locations of the solutions can be seen in the following graph.

Find all the solutions of the following equation: $\sin^2x+\left(\frac{\sqrt{3}-1}{2}\right)\cos x=1-\frac{\sqrt{3}}{4}$`s``i``n`2`x`+(√3−12)`c``o``s``x`=1−√34.

This looks difficult but we take it in small steps.

We use the Pythagorean identity to obtain $1-\cos^2x+\left(\frac{\sqrt{3}-1}{2}\right)\cos x=1-\frac{\sqrt{3}}{4}$1−`c``o``s`2`x`+(√3−12)`c``o``s``x`=1−√34.

Then, after collecting like terms and reversing the signs, $\cos^2x+\left(\frac{1-\sqrt{3}}{2}\right)\cos x=\frac{\sqrt{3}}{4}$`c``o``s`2`x`+(1−√32)`c``o``s``x`=√34. We see that the equation is a quadratic equation in $\cos x$`c``o``s``x`. We could resort immediately to the quadratic formula, or perhaps rewrite it in factorised form as

$\left(\cos x+\frac{1}{2}\right)\left(\cos x-\frac{\sqrt{3}}{2}\right)=0$(`c``o``s``x`+12)(`c``o``s``x`−√32)=0

From here, we see that $\cos x=-\frac{1}{2}$`c``o``s``x`=−12 and $\cos x=\frac{\sqrt{3}}{2}$`c``o``s``x`=√32 are solutions to the quadratic equation.

Hence, in the second and third quadrants, $x=\pi-\frac{\pi}{3}$`x`=π−π3 and $x=\pi+\frac{\pi}{3}$`x`=π+π3, while in the first and fourth quadrants, $x=\frac{\pi}{6}$`x`=π6 and $x=2\pi-\frac{\pi}{6}$`x`=2π−π6.

Putting this together, we can write the general solution as

$x=\frac{2\pi}{3}+2n\pi$`x`=2π3+2`n`π or $x=\frac{4\pi}{3}+2n\pi$`x`=4π3+2`n`π or $x=\frac{\pi}{6}+2n\pi$`x`=π6+2`n`π or $x=\frac{11\pi}{6}+2n\pi$`x`=11π6+2`n`π.

The right- and left-hand sides of the equation have been graphed. The solutions are where the graphs intersect.

Find the general solution of $\tan2x=\cot x$`t``a``n`2`x`=`c``o``t``x` in radians.

You may use $n$`n` to represent an integer.

Form and use trigonometric, polynomial, and other non-linear equations

Apply trigonometric methods in solving problems