Trigonometric Equations, Summaries of Pre-Calculus

How to solve trigonometric equations, which are equations involving trigonometric functions that are satisfied only by some values, or possibly no values, of the variable. It provides examples of how to find all the solutions of a trigonometric equation and how to use a calculator to solve them.

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Math Analysis Precalculus, Sullivan 10th Edition
Ch7Sec3Day1 2/23/2020
1
Section 7.3 Trigonometric Equations Day 1
In this section you will look at trigonometric equations, which are equations involving trigonometric functions that
are satisfied only by some values, or possibly no values, of the variable. The values that satisfy the equation are
called solutions of the equation.
Unless the domain of the variable is restricted, you need to find all the solutions of a trigonometric equation.
To find all the solutions, first find solutions over an interval whose length equals the period of the function and
then add multiples of that period to the solutions found.
Example 1: Solve the equation
.
2
3
cos
The period of the cosine function is
.2
In the interval
,2,0
there are two angles for which
6
:
2
3
cos
and
6
11
The equation has an infinite number of solutions due to the periodicity of the cosine function.
Because the cosine has period
,2
all the solutions of
2
3
cos
may be given by
and
,k2
6
11
where k is any integer.
To check your solution, you may graph
xcosy1
and
23
y,
2
in radian mode. The solutions
are where the graphs intersect.
Example 2: Solve the equation
.20,1)2sin(
The period of the sine function is
.2
In the interval
,2,0
the sine function has the value 1 only at an angle of
.
2
So,
,k2
2
2
k any integer.
,k
4
k any integer.
But the domain is restricted to
,20
so it follows that
4
and
4
5
are the only
solutions.
To check your solution, you may graph
)x2sin(y1
and
2
y 1,
in radian mode. The solutions
are where the graphs intersect.
In using a calculator to solve trigonometric equations, remember that the calculator supplies an angle only within
the restrictions of the definition of the inverse trigonometric function. To find the remaining solutions, you must
identify other quadrants, if any, in which the angle may be located.
Example 3: Use a calculator to solve
4.0cos
for
.20
)4.0(cos 1
Your calculator gives
,159279.1
an angle in the first quadrant. But, cosine is also positive in
quadrant IV.
The angle
159279.122
123906.5
is the angle in quadrant IV where
.4.0cos
Thus, the solution to
4.0cos
for
20
is
12.5and16.1
radians.
pf2

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Math Analysis – Precalculus, Sullivan 10th^ Edition

Ch7Sec3Day1 1 2/23/

Section 7.3 – Trigonometric Equations – Day 1

In this section you will look at trigonometric equations , which are equations involving trigonometric functions that

are satisfied only by some values, or possibly no values, of the variable. The values that satisfy the equation are

called solutions of the equation.

Unless the domain of the variable is restricted, you need to find all the solutions of a trigonometric equation.

To find all the solutions, first find solutions over an interval whose length equals the period of the function and

then add multiples of that period to the solutions found.

Example 1: Solve the equation. 2

cos 

The period of the cosine function is 2 .

In the interval  0 , 2 ,there are two angles for which

cos

  and 6

The equation has an infinite number of solutions due to the periodicity of the cosine function.

Because the cosine has period 2 ,all the solutions of 2

cos  may be given by

  2 k 6

and 2 k , 6

 where k is any integer.

To check your solution, you may graph y 1 cosxand 2

y , 2

 in radian mode. The solutions

are where the graphs intersect.

Example 2: Solve the equationsin( 2 ) 1 , 0  2 .

The period of the sine function is 2 .

In the interval  0 , 2 ,the sine function has the value 1 only at an angle of. 2

So, 2 k , 2

 k any integer.

k , 4

 k any integer.

But the domain is restricted to 0  2 ,so it follows that 4

 and 4

 are the only

solutions.

To check your solution, you may graph y 1 sin( 2 x)and y 2 1,in radian mode. The solutions

are where the graphs intersect.

In using a calculator to solve trigonometric equations, remember that the calculator supplies an angle only within

the restrictions of the definition of the inverse trigonometric function. To find the remaining solutions, you must

identify other quadrants, if any, in which the angle may be located.

Example 3: Use a calculator to solve cos^ ^0.^4 for^0 ^2 .

cos ( 0. 4 )

 1  

Your calculator gives ^1.^159279 ,an angle in the first quadrant. But, cosine is also positive in

quadrant IV.

The angle 2    2  1. 159279

 5. 123906 is the angle in quadrant^ IV^ wherecos^ ^0.^4.

Thus, the solution to cos  0. 4 for 0   2  is  1. 16 and  5. 12 radians.

Math Analysis – Precalculus, Sullivan 10th^ Edition

Ch7Sec3Day1 2 2/23/

General Solutions

(General Solutions)

Section 7.3 – Trigonometric Equations – Day 1 (continued)

Many trigonometric equations can be solved by factoring or by applying the quadratic formula. When a

trigonometric equation contains more than one trigonometric function, identities may be used to create an

equivalent equation that contains only one trigonometric function. If you square both sides of an equation,

remember to check your solutions because extraneous solutions may be introduced.

Example 4: Solve the equation 3 cos 3 2 sin , 0 2.

2     

2 3 cos 3 2 sin

3 cos 3 2 ( 1 cos )

2     Use the Pythagorean Identity

2 3 cos 3 2 2 cos

2 cos 3 cos 1 0

2    This is a quadratic equation incos

( 2 cos 1 )(cos 1 ) (^0) Factor

 2 cos 1  (^0) or cos  1  (^0) By the Zero Product Property

2 cos 1 cos  1

cos

  2 k, k any integer

2 k , 3

 k any integer

and

2 k , 3

 k any integer

On the interval ^0 ,^2 ,the solutions are

, ,and 3

Example 5: Example 6:

Solve the equationcos sin 4.

2   Solve the equation:

sin( 2 ) , 0 2. 2

cos sin 4

2  

sin( 2 ) 2

 1 sin  sin 4

2

    Pythagorean Identity Sine is positive in quadrants I and II, so

sin sin 3 0

2      

  2 k 6

2 and 2 k , 6

 k any integer

sin sin 3 0

2     

  k 12

and k , 12

 k any integer

This is a quadratic equation insin .

Discriminantb 4 ac ( 1 ) 4 ( 1 )( 3 )

2 2    

 1  12 On the interval ^0 , 2 ,the solutions are

, and 12

 There is no real solution.

All material has been taken from Precalculus, by M. Sullivan, 10th^ Edition