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5.3 Solving Trig equationsPowerPoint Presentation

5.3 Solving Trig equations

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5.3 Solving Trig equations

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5.3 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:
- Isolate the trig function
- Find the reference angle
- Put the reference angle in the proper quadrant(s)
- 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) Put the reference angle in the proper quadrant(s)

I =

IV =

Cos x = ½

4) Create a formula if necessary

x =

x =

- 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.:

III:

Iv:

- Find the solutions in the interval [0, 2π) for the following equation:
Tan²x – 3 = 0

Tan²x = 3

Tan x =

Tan x =

Ref. Angle:

Quad.:

I:

III:

IV:

II:

x =

- Solve the following equations for all real values of x.
- Sin x + = - Sin x
- 3Tan² x – 1 = 0
- Cot x Cos² x = 2 Cot x

- Find all solutions to the following equation:
Sin x + = - Sin x

2 Sin x = -

x =

Sin x = -

x =

3Tan² x – 1 = 0

x =

Tan² x =

x =

Tan x =

x =

x =

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

Cos x = 0

Cos² x – 2 = 0

x =

Cos x =

No Solution

x =

5.3 Solving Trig equations

- Find all solutions to the following equation.
4 Tan²x – 4 = 0

x =

Tan²x = 1

x =

Tan x = ±1

Ref. Angle =

- Equations of the Quadratic Type
- Many trig equations are of the quadratic type:
- 2Sin²x – Sin x – 1 = 0
- 2Cos²x + 3Sin x – 3 = 0

- Solve the following on the interval [0, 2π)
2Cos²x + Cos x – 1 = 0

2x² + x - 1

If possible, factor the equation into two binomials.

(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

Cos x = -1

Ref. Angle:

x =

Quad:

I, IV

x =

- Solve the following on the interval [0, 2π)
2Sin²x - Sin x – 1 = 0

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

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

Sin x = - ½

Sin x = 1

Ref. Angle:

x =

Quad:

III, IV

x =

- Solve the following on the interval [0, 2π)
2Cos²x + 3Sin x – 3 = 0

Convert all expressions to one trig function

2 (1 – Sin²x) + 3Sin x – 3 = 0

2 – 2Sin²x + 3Sin x – 3 = 0

0 = 2Sin²x – 3Sin x + 1

0 = 2Sin²x – 3Sin x + 1

0 = (2Sin x – 1) (Sin x – 1)

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

Sin x = ½

Sin x = 1

Ref. Angle:

x =

Quad:

I, II

x =

- Solve the following on the interval [0, 2π)
2Sin²x + 3Cos x – 3 = 0

Convert all expressions to one trig function

2 (1 – Cos²x) + 3Cos x – 3 = 0

2 – 2Cos²x + 3Cos x – 3 = 0

0 = 2Cos²x – 3Cos x + 1

0 = 2Cos²x – 3Cos x + 1

0 = (2Cos x – 1) (Cos x – 1)

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

Cos x = ½

Cos x = 1

Ref. Angle:

x =

Quad:

I, IV

x =

- 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 =

(Cos x + 1)² = Sin² x

Cos x + 1 = Sin x

Cos²x + 2Cos x + 1 = 1 – Cos² x

2Cos² x + 2 Cos x = 0

Cos x (2 Cos x + 2) = 0

Cos x = 0

Cos x = - 1

x =

x =

Cos x + 1 = Sin x

x =

5.3 Solving Trig equations

- 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 = 0

2Sin 2x = -1

Sin 2x = - ½

2x =

2x =

x =

x =

Redundant

Answer

- Solve the following equations for all values of x.
- 2Cos 3x – 1 = 0
- Cot (x/2) + 1 = 0

2Cos 3x - 1 = 0

2Cos 3x = 1

Cos 3x = ½

3x =

3x =

x =

x =

- 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

- Solving basic trig equations

Find all solutions to the following equation:

Sec²x – 3Sec x – 10 = 0

(Sec x + 2) (Sec x – 5) = 0

Sec x + 2 = 0 Sec x – 5 = 0

Sec x = 5

Sec x = -2

Cos x =

Cos x = - ½

x =

x =

- One of the following equations has solutions and the other two do not. Which equations do not have solutions.
- Sin²x – 5Sin x + 6 = 0
- Sin²x – 4Sin x + 6 = 0
- 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.

- 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 = 3

Tan x = -1

Tan x = -1

Tan x = 3

x = ArcTan 3

ref. angle:

71.6º

I, III

Quad:

x =

71.6º,

251.6º