5.3 Solving Trig equations

5.3 SOLVING TRIG EQUATIONS

SOLVING TRIG EQUATIONS

Solve the following equation for x:

Sin x = ½


In this section, we will be solving various types of trig equations

You will need to use all the procedures learned last year in Algebra II

All of your answers should be angles.

Note the difference between finding all solutions and finding all solutions in the domain [0, 2π)


Guidelines to solving trig equations:

1) Isolate the trig function

2) Find the reference angle

3) Put the reference angle in the proper quadrant(s)

4) Create a formula for all possible answers (if necessary)


1- 2 Cos x = 0

1) Isolate the trig function

1- 2 Cos x = 0 + 2 Cos x = + 2 Cos x

1= 2 Cos x 2 2

Cos x = ½


Cos x = ½

2) Find the reference angle

x = 3

3) Put the reference angle in the proper quadrant(s)

I = 3

IV =

3

5


Cos x = ½

4) Create a formula if necessary

x = 3

x = 3

5

2n

2n


Find all solutions to the following equation:

Sin x + 1 = - Sin x+ Sin x + Sin x

→ 2 Sin x + 1 = 0- 1 - 1

→ 2 Sin x = -1→ Sin x = - ½


Sin x = - ½

Ref. Angle: Quad.:6

IV III,

III:6

7

Iv:6

11

2n

2n


Find the solutions in the interval [0, 2π) for the following equation:

Tan²x – 3 = 0

Tan²x = 3Tan x = 3


Tan x = 3

Ref. Angle: Quad.:3

IV III, II, I,

I:3

II:

3

2 III:3

4IV:

3

5

x = ,3

,

3

2,

3

43

5


Solve the following equations for all real values of x.

a) Sin x + = - Sin x

b) 3Tan² x – 1 = 0

c) Cot x Cos² x = 2 Cot x

2



Sin x + = - Sin x

2 Sin x = -

Sin x = -

2

2

2

2

x = 4

5

x = 4

7

2n

2n


3Tan² x – 1 = 0

Tan² x = 3

1

Tan x = 3

1

x = 6

6

5x =

6

7x =

6

11x =

2n

2n

2n

2n


Cot x Cos² x = 2 Cot x

Cot x Cos² x – 2 Cot x = 0

Cot x (Cos² x – 2) = 0

Cot x = 0 Cos² x – 2 = 0

Cos² x – 2 = 0 Cos x = 2

Cos x = 0 x =

2

x = 2

3

2n

2n No Solution


2



Find all solutions to the following equation.

4 Tan²x – 4 = 0 Tan²x = 1 Tan x = ±1 Ref. Angle =

4

x = 4

4

3x =

n

n


Equations of the Quadratic Type

Many trig equations are of the quadratic type:

2Sin²x – Sin x – 1 = 0 2Cos²x + 3Sin x – 3 = 0

To solve such equations, factor the quadratic or, if that is not possible, use the quadratic formula


Solve the following on the interval [0, 2π)

2Cos²x + Cos x – 1 = 0

If possible, factor the equation into two binomials.

2x² + x - 1

(2Cos x – 1) (Cos x + 1) = 0

Now set each factor equal to zero


2Cos x – 1 = 0 Cos x + 1 = 0

Cos x = ½

Ref. Angle: 3

Quad: I, IV

x = ,3

3

5

Cos x = -1

x =



2Sin²x - Sin x – 1 = 0

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


2Sin x + 1 = 0 Sin x - 1 = 0

Sin x = - ½

Ref. Angle: 6

Quad: III, IV

x = ,6

76

11

Sin x = 1

x = 2



2Cos²x + 3Sin x – 3 = 0

Convert all expressions to one trig function

2 (1 – Sin²x) + 3Sin x – 3 = 02 – 2Sin²x + 3Sin x – 3 = 0

0 = 2Sin²x – 3Sin x + 1


2Sin x - 1 = 0 Sin x - 1 = 0

Sin x = ½

Ref. Angle: 6

Quad: I, II

x = ,6

6

5

Sin x = 1

x = 2

0 = 2Sin²x – 3Sin x + 1 0 = (2Sin x – 1) (Sin x – 1)



2Sin²x + 3Cos x – 3 = 0

Convert all expressions to one trig function

2 (1 – Cos²x) + 3Cos x – 3 = 02 – 2Cos²x + 3Cos x – 3 = 0

0 = 2Cos²x – 3Cos x + 1


2Cos x - 1 = 0 Cos x - 1 = 0

Cos x = ½

Ref. Angle: 3

Quad: I, IV

x = ,3

3

5

Cos x = 1

x = 0

0 = 2Cos²x – 3Cos x + 1 0 = (2Cos x – 1) (Cos x – 1)


The last type of quadratic equation would be a problem such as:

Sec x + 1 = Tan x

What do these two trig functions have in common?

When you have two trig functions that are related

through a Pythagorean Identity, you can square

both sides.

( )² ²


(Sec x + 1)² = Tan²x

Sec²x + 2Sec x + 1 = Sec²x - 1

2 Sec x + 1 = -1 Sec x = -1

Cos x = -1

x = When you have a problem that requires you to square both sides, you must check your answer when you are done!


Sec x + 1 = Tan x x =

1 1 0


Cos x + 1 = Sin x

Cos²x + 2Cos x + 1 = 1 – Cos² x

2Cos² x + 2 Cos x = 0Cos x (2 Cos x + 2) = 0

Cos x = 0

x = ,2

(Cos x + 1)² = Sin² x

Cos x = - 1

2

3x =


Cos x + 1 = Sin x

x = ,2

,

2

3 2

Sin 1 2

Cos

1 1 0

2

3Sin 1

2

3 Cos

1- 1 0

Sin 1 Cos

0 1 1-



Equations involving multiply angles

Solve the equation for the angle as your normally would

Then divide by the leading coefficient


Solve the following trig equation for all values of x.

2Sin 2x + 1 = 02Sin 2x = -1

Sin 2x = - ½

2x = 2n

6

7 2x =

2n 6

11

x = n

12

7 x =

n 12

11


0 3 2

xTan 3

1- 2

xTan

2

x n

4

3

2

x n

4

7

2

x 2n

2

3

2

x 2n

2

7

Redundant Answer


Solve the following equations for all values of x.

a) 2Cos 3x – 1 = 0

b) Cot (x/2) + 1 = 0


2Cos 3x - 1 = 02Cos 3x = 1

Cos 3x = ½

3x = 2n

3 3x =

2n 3

5

x = 3

2n

9

x =

3

2n

9

5


0 1 2

xCot

1- 2

xCot

2

x n

4

3

2

x 2n

2

3


Topics covered in this section:

Solving basic trig equations Finding solutions in [0, 2π) Find all solutions

Solving quadratic equations Squaring both sides and solving Solving multiple angle equations Using inverse functions to generate answers



Sec²x – 3Sec x – 10 = 0(Sec x + 2) (Sec x – 5) = 0

Sec x + 2 = 0 Sec x – 5 = 0Sec x = -2Cos x = - ½

x = 2n 3

2

2n 3

4

Sec x = 5Cos x =

5

1

x =

5

1Cos 1-


One of the following equations has solutions and the other two do not. Which equations do not have solutions.

a) Sin²x – 5Sin x + 6 = 0b) Sin²x – 4Sin x + 6 = 0c) Sin²x – 5Sin x – 6 = 0

Find conditions involving constants b and c that

will guarantee the equation Sin²x + bSin x + c = 0

has at least one solution.

SOLVING TRIG FUNCTIONS

Find all solutions of the following equation in the interval [0, 2π)

Sec²x – 2 Tan x = 4

1 + Tan²x – 2Tan x – 4 = 0

Tan²x – 2Tan x – 3 = 0

(Tan x + 1) (Tan x – 3) = 0

Tan x = -1 Tan x = 3

SOLVING TRIG FUNCTIONS

Tan x = -1 Tan x = 3

4

7 ,

4

3 X

x = ArcTan 3

ref. angle: 71.6º

Quad: I, III

x = 71.6º, 251.6º

5.3 Solving Trig equations

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