Trigonometric equations

TRIGONOMETRICEQUATIONS

Firstly recall the graphs of y = sin x, y = cos x and y = tan x.

y = sin x

y = cos x

y = tan x

Note: sin x is positivefor 0º < x < 180º, and negative for 180º < x < 360º.

Note: cos x is positivefor 0º ≤ x < 90º, andfor 270º < x ≤ 360º

and negative for 90º < x < 270º.

Note: tan x is positivefor 0° < x < 90º, andfor 180º < x < 270º

and negative for 90º < x < 180º. andfor 270º < x < 360º.

We can summarise the information in the following diagram:

0º, 360º180º

90º

270º

sin +vecos +vetan +ve

sin +vecos –vetan –ve

sin –vecos –vetan +ve

sin –vecos +vetan –ve

This can be simplified to show just the positive ratios:

0º, 360º180º

90º

270º

all +vesin +ve

tan +ve cos +ve

or just:AS

T C

N.B. You may find a mnemonic will help you memorise the positions.

Solving Trigonometric Equations

Quadrant IQuadrant II

Quadrant III Quadrant IV

Cosine

AllSine

Tangent

CAST Rule

Find the measure of 0 ≤ < 3600

a) cos = -0.6691

The reference angle is 480. The angle is foundin Quadrants II and III.

1320 and 2280

b) tan = 1.2435The reference angle is 510. The angle is foundin Quadrants I and III.

510 and 2310

5.2.2

Solving Trigonometric Equationswith a Given Interval

Example : Solve sin x = –0.5; –360º < x ≤ 360º

The sine of an angle is negative in the

α = sin–1 0.5 = 30º

Hence the solutions in the given range are:

x =

In problems where the angle is not simply x, the given rangewill need to be adjusted.

–150°, –30°, 210°, 330°

N.B. The negative is ignoredto find the acute angle.

3rd and 4th quadrants.

The positive ratio diagram can be used to solve trigonometric equations:

Example : Solve cos x = 0.5; 0º ≤ x < 360º

Using: AS

T C

We can see that the cosine of anangle is positive in the two quadrantson the right, i.e. the 1st and 4th

quadrants.

α = cos–1 0.5 = 60º

Hence the solutions in the given range are:

x = so x = 60º, 300º

60°, 360 – 60°

Example 3: Solve 3 + 5 tan 2x = 0; 0º ≤ x ≤ 360º.

Firstly we need to make tan 2x the subject of the equation:

The tangent of an angle is negative in the

53tan α 1 = 30.96°

The range must be adjusted for the angle 2x. i.e. 0° ≤ 2x ≤ 720°.

Hence: 2x =

x = 74.5°, 164.5°, 254.5°, 344.5°.

149.04°, 329.04°, 509.04°, 689.04°.

2nd and 4th quadrants.

tan 2x = –35

Solve Trigonometric Equations

• Trig equations are usually solved for a values of the variable between 0 degrees and 360 degrees.

• But there are solutions outside that interval.

• These other solutions differ by integral multiples of the period of the function

-1

x

y

1-19π

6-11π

6 -7π

6 π

6 5π

6 13π

6 17π

6 25π

6

-π-2π-3π π 2π 3π 4π

All the solutions for x can be expressed in the form of a general solution.

x = is one of infinitely many solutions of y = sin x.π 6

x = + 2k π and x = 5 + 2k π (k = 0, ±1, ± 2, ± 3, ). 6π π

6

sin x = is a trigonometric equation.2

1

y =2

1

Example: Solve tan x = 1.The graph of y = 1 intersects the graph of y = tan x infinitely many times.

General solution: x = + kπ for k any integer. π 4

Points of intersection are at x = and every multiple of π added or subtracted from .π

4

π 4

y

2

x-π π 2π 3π

x = -3π2

y = tan(x)

x = -π2

x = π x = 3π2

x = 5π2

y = 1

- π – 2π4

- π – π4

π4

π + π4

π + 2π4

π + 3π4

1-π 4

2

Example:

Solve the equation 3sin x + = sin x for ≤ x ≤ .π 2

π 2

2sin x + = 0 Collect like terms.2

3sin x sin x + = 02

3sin x + = sin x2

sin x = 2

2

1 x

y

y = -2

2

x = is the only solution in the interval ≤ x ≤ . π 2

π 2

π 4

Solving Second Degree (Quadratic) Trigonometric Equations

Notes

Solving Second Degree Equations

Example: Solve. 0° ≤ x ≤ 360° tan2x – 3tanx – 4 = 0

let x = tan x

x2 – 3x – 4 = 0(x – 4)(x + 1) = 0

x – 4 = 0 x + 1 = 0x = 4 x = -1

tan x = 4 tan x = -1

x = 76° x = 135°x = 256° x = 315°


Example: Solve. 0° ≤ x ≤ 360°

3cos2x – 5cosx = 4

let x = cos x

3x2 – 5x – 4 = 0

a

acbbx

2

42

0.59

2.266735

32

434255

x

x

x

x


x = 2.26x = -0.59

cos x = 2.26 cos x = -0.59

x = Ø x = 126°

x = 234°


Example: Solve. 0° ≤ x ≤ 360° 2cos2x = cosx

let x = cos x

2x2 = x2x2 – x = 0x(2x – 1) = 0

x = 0 2x – 1 = 0x = .5

cos x = 0 cos x = .5

x = 90° x = 270°x = 270°

Solving Second Degree EquationsExample: Solve. 0° ≤ x ≤ 360°

3sinx + 4 = 1/sinx

let x = sin x

3x + 4 = 1/x

x(3x + 4) = x(1/x) 3x2 + 4x = 1(3x – 1)(x + 1) = 0


3x - 1= 0 x + 1 = 0

x = .333 x = -1

sin x = .333 sin x = -1

x = 19° x = 270°

x = 161°


Example: Solve. 0° ≤ x ≤ 360°

2cos2x – sinx = 1

Use the identity cos2x = 1 – sin2x

2 (1 – sin2x) – sinx = 1

2 – 2sin2x – sinx = 1

– 2sin2x – sinx + 1 = 0

2sin2x + sinx - 1 = 0


2sin2x + sinx - 1 = 0

let x = sin x

2x2 + x - 1 = 0(2x - 1)(x + 1) = 0

2x - 1 = 0 x + 1 = 0x = ½ x = -1

sin x = ½ sin x = -1x = 30° x = 270°x = 150°

The trigonometric equation 2 sin2 + 3 sin + 1 = 0 is quadratic in form.

2 sin2 + 3 sin + 1 = 0 implies that

x

= -π + 2kπ, from sin = -1

(2 sin + 1)(sin + 1) = 0.

Therefore, 2 sin + 1 = 0 or sin + 1 = 0.

Solutions: = - + 2kπ and = + 2kπ, from sin = -π

67π 6

1 2

It follows that sin = - or sin = -1.1 2

8 sin = 3(1 sin2 ) Use the Pythagorean Identity.

Rewrite the equation in terms of only one trigonometric function.

Example: Solve 8 sin = 3 cos2 with in the

interval [0, 2π].

3 sin2 + 8 sin 3 = 0. A “quadratic” equation with sin x as the variable

Therefore, 3 sin 1 = 0 or sin + 3 = 0

(3 sin 1)(sin + 3) = 0 Factor.

s

Solutions: sin = or sin = -31 3

= sin1( ) = 0.3398 and = π sin1( ) = 2.8107.1 3

1 3

Solve: 5cos2 + cos – 3 = 0 for 0 ≤ ≤ π.

This is the range of the inverse cosine function.

The solutions are: = cos 1(0.6810249 ) = 0.8216349 and = cos 1(0.8810249) = 2.6488206

Therefore, cos = 0.6810249 or –0.8810249.

Use the calculator to find values of in 0 ≤ ≤ π.

The equation is quadratic. Let y = cos and solve 5y2 + y 3 = 0.

y = (-1 ± ) = 0.6810249 or -0.881024961 10

Example: Solve 6 sin2 x + sin x – 1 = 0; 0º ≤ x < 360º

A quadratic equation!

It may help to abbreviate sin x with s:

i.e. 6s2 + s – 1 = 0

Factorising this: (3s – 1)(2s + 1)= 0

1 1sin or sin

3 2 x x ∴

α = 19.47º α = 30º

So, x = (To nearest 0.1º.)19.5°, 160.5°, 210°, 330°.

Summary of key points:

To solve a Trigonometric Equation:

• Re-arrange the equation to make sin, cos or tan of some angle the subject.

• Locate the quadrants in which the ratio is positive, or negative as required.

AST C

Using:

• Adjust the range for the given angle.

• Read off all the solutions within the range.

This PowerPoint produced by R. Collins; © ZigZag Education 2008–2010

Trigonometric equations

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