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OBJECTIVES:

1. Basic Trigonometric Equations

2. Solving Trigonometric Equations by Factoring

1. Basic Trigonometric Equations

2. Solving Trigonometric Equations by Factoring

1.
Analytic Trigonometry

Copyright © Cengage Learning. All rights reserved.

Copyright © Cengage Learning. All rights reserved.

2.
Basic Trigonometric

Equations

Copyright © Cengage Learning. All rights reserved.

Equations

Copyright © Cengage Learning. All rights reserved.

3.
► Basic Trigonometric Equations

► Solving Trigonometric Equations by

Factoring

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► Solving Trigonometric Equations by

Factoring

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4.
Basic Trigonometric Equations

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5.
Basic Trigonometric Equations

An equation that contains trigonometric functions is called a

trigonometric equation. For example, the following are

trigonometric equations:

sin2 + cos2 = 1 2 sin – 1 = 0 tan 2 – 1 = 0

The first equation is an identity—that is, it is true for every

value of the variable . The other two equations are true

only for certain values of .

To solve a trigonometric equation, we find all the values of

the variable that make the equation true.

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An equation that contains trigonometric functions is called a

trigonometric equation. For example, the following are

trigonometric equations:

sin2 + cos2 = 1 2 sin – 1 = 0 tan 2 – 1 = 0

The first equation is an identity—that is, it is true for every

value of the variable . The other two equations are true

only for certain values of .

To solve a trigonometric equation, we find all the values of

the variable that make the equation true.

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6.
Basic Trigonometric Equations

Solving any trigonometric equation always reduces to

solving a basic trigonometric equation—an equation of

the form Trig( ) = c, where Trig is a trigonometric function

and c is a constant. For example:

2 sin – 1 = 0

In the next example we solve such basic equations.

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Solving any trigonometric equation always reduces to

solving a basic trigonometric equation—an equation of

the form Trig( ) = c, where Trig is a trigonometric function

and c is a constant. For example:

2 sin – 1 = 0

In the next example we solve such basic equations.

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7.
Example – Solving Trigonometric Equations

Find all solutions of the equation.

(a)2 sin – 1 = 0

(a) We start by isolating sin :

2 sin – 1 = 0 Given equation

Add 1

2 sin = 1

Divide by 2

sin =

Now use the table at the bottom of the identity formula sheet and look to see

Where sin equals 1/2

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Find all solutions of the equation.

(a)2 sin – 1 = 0

(a) We start by isolating sin :

2 sin – 1 = 0 Given equation

Add 1

2 sin = 1

Divide by 2

sin =

Now use the table at the bottom of the identity formula sheet and look to see

Where sin equals 1/2

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8.
Example – Solution cont’d

This time we must consider that since sine is positive in the

first and second quadrants, the solutions are

= + 2k = + 2k

where k is any integer.

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This time we must consider that since sine is positive in the

first and second quadrants, the solutions are

= + 2k = + 2k

where k is any integer.

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9.
Example – Solution cont’d

(b) tan2 – 3 = 0

We start by isolating tan :

tan2 – 3 = 0 Given equation

tan2 = 3 Add 3

tan Take the square root

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(b) tan2 – 3 = 0

We start by isolating tan :

tan2 – 3 = 0 Given equation

tan2 = 3 Add 3

tan Take the square root

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Example – Solution cont’d

Because tangent is positive in both QI & QIII and negative

in QII & QIV, first find one value for the positive tan and

then one value for the negative tan and finally to get all

solutions, we add integer multiples of or 180 º to these

= + k =2 + k

OR in Degrees:

= 60º + 180k = 120º + 180k

where k is any integer.

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Because tangent is positive in both QI & QIII and negative

in QII & QIV, first find one value for the positive tan and

then one value for the negative tan and finally to get all

solutions, we add integer multiples of or 180 º to these

= + k =2 + k

OR in Degrees:

= 60º + 180k = 120º + 180k

where k is any integer.

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11.
Solving Trigonometric Equations

by Factoring

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by Factoring

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12.
Solving Trigonometric Equations by Factoring

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13.
Example – A Trigonometric Equation of Quadratic Type

Now solve the equation 2 cos2 – 7 cos + 3 = 0.

We factor the left-hand side of the equation.

2 cos2 – 7 cos + 3 = 0 Given equation

(2 cos – 1)(cos – 3) = 0 Factor

2 cos – 1 = 0 or cos – 3 = 0 Set each factor equal to 0

cos = or cos = 3 Solve for cos

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Now solve the equation 2 cos2 – 7 cos + 3 = 0.

We factor the left-hand side of the equation.

2 cos2 – 7 cos + 3 = 0 Given equation

(2 cos – 1)(cos – 3) = 0 Factor

2 cos – 1 = 0 or cos – 3 = 0 Set each factor equal to 0

cos = or cos = 3 Solve for cos

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14.
Example – Solution cont’d

Because cosine has period 2, we first find the solutions in

the interval [0, 2). For the first equation the solutions are

= /3 and = 5 /3 (see Figure 7 and your formula

Figure 7

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Because cosine has period 2, we first find the solutions in

the interval [0, 2). For the first equation the solutions are

= /3 and = 5 /3 (see Figure 7 and your formula

Figure 7

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15.
Example – Solution cont’d

The second equation has no solution because cos is

never greater than 1.

Thus the solutions are

= + 2k = + 2k

where k is any integer.

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The second equation has no solution because cos is

never greater than 1.

Thus the solutions are

= + 2k = + 2k

where k is any integer.

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16.
Example – Solving a Trigonometric Equation by Factoring

Given equation

Factor

Set each factor equal to 0

Solve for sin

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Given equation

Factor

Set each factor equal to 0

Solve for sin

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17.
Example – Solution cont’d

Because sine and cosine have period 2 or 360º we first

find the solutions of these equations in an interval of length

For the first equation cos = 0 ,

the solutions in the interval [0, 360º) are = 90º and

= 270º. To solve the second equation, we take sin–1 of

each side:

sin = –0.80

Second equation

= sin–1(–0.80) Take sin–1 of each side with your

calculator let’s use degree mode

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Because sine and cosine have period 2 or 360º we first

find the solutions of these equations in an interval of length

For the first equation cos = 0 ,

the solutions in the interval [0, 360º) are = 90º and

= 270º. To solve the second equation, we take sin–1 of

each side:

sin = –0.80

Second equation

= sin–1(–0.80) Take sin–1 of each side with your

calculator let’s use degree mode

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18.
Example – Solution cont’d

- 53.13 º Calculator (in degree mode)

So the solutions in an interval of length 360ºare = - 53.13 º

and then since sine is also negative in the third quadrant,

= 180 + 53.13 or 233.13º

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- 53.13 º Calculator (in degree mode)

So the solutions in an interval of length 360ºare = - 53.13 º

and then since sine is also negative in the third quadrant,

= 180 + 53.13 or 233.13º

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19.
Example – Solution cont’d

We get all the solutions of the equation by adding integer

multiples of 360º to these solutions.

= 90 + 360k, = 270 +360k

– 53.13 + 360k 233.13 + 360k

where k is any integer.

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We get all the solutions of the equation by adding integer

multiples of 360º to these solutions.

= 90 + 360k, = 270 +360k

– 53.13 + 360k 233.13 + 360k

where k is any integer.

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