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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 = Â½

- 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, 2Ï€)

- 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 = Â½

Cos x = Â½

2) Find the reference angle

x =

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

I =

IV =

Cos x = Â½

4) Create a formula if necessary

x =

x =

- Find all solutions to the following equation:
Sin x + 1 = - Sin x

+ Sin x + Sin x

â†’ 2 Sin x + 1 = 0

- 1 - 1

â†’ 2 Sin x = -1

â†’ Sin x = - Â½

Sin x = - Â½

Ref. Angle:

Quad.:

III:

Iv:

- Find the solutions in the interval [0, 2Ï€) for the following equation:
TanÂ²x â€“ 3 = 0

TanÂ²x = 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
- 3TanÂ² x â€“ 1 = 0
- Cot x CosÂ² x = 2 Cot x

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

2 Sin x = -

x =

Sin x = -

x =

3TanÂ² x â€“ 1 = 0

x =

TanÂ² x =

x =

Tan x =

x =

x =

Cot x CosÂ² x = 2 Cot x

Cot x CosÂ² x â€“ 2 Cot x = 0

Cot x (CosÂ² x â€“ 2) = 0

Cot x = 0

CosÂ² x â€“ 2 = 0

Cos x = 0

CosÂ² x â€“ 2 = 0

x =

Cos x =

No Solution

x =

5.3 Solving Trig equations

- Find all solutions to the following equation.
4 TanÂ²x â€“ 4 = 0

x =

TanÂ²x = 1

x =

Tan x = Â±1

Ref. Angle =

- Equations of the Quadratic Type
- Many trig equations are of the quadratic type:
- 2SinÂ²x â€“ Sin x â€“ 1 = 0
- 2CosÂ²x + 3Sin x â€“ 3 = 0

- Solve the following on the interval [0, 2Ï€)
2CosÂ²x + Cos x â€“ 1 = 0

2xÂ² + x - 1

If possible, factor the equation into two binomials.

(2Cos x â€“ 1) (Cos x + 1) = 0

Now set each factor equal to zero

2Cos x â€“ 1 = 0 Cos x + 1 = 0

Cos x = Â½

Cos x = -1

Ref. Angle:

x =

Quad:

I, IV

x =

- Solve the following on the interval [0, 2Ï€)
2SinÂ²x - Sin x â€“ 1 = 0

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

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

Sin x = - Â½

Sin x = 1

Ref. Angle:

x =

Quad:

III, IV

x =

- Solve the following on the interval [0, 2Ï€)
2CosÂ²x + 3Sin x â€“ 3 = 0

Convert all expressions to one trig function

2 (1 â€“ SinÂ²x) + 3Sin x â€“ 3 = 0

2 â€“ 2SinÂ²x + 3Sin x â€“ 3 = 0

0 = 2SinÂ²x â€“ 3Sin x + 1

0 = 2SinÂ²x â€“ 3Sin x + 1

0 = (2Sin x â€“ 1) (Sin x â€“ 1)

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

Sin x = Â½

Sin x = 1

Ref. Angle:

x =

Quad:

I, II

x =

- Solve the following on the interval [0, 2Ï€)
2SinÂ²x + 3Cos x â€“ 3 = 0

Convert all expressions to one trig function

2 (1 â€“ CosÂ²x) + 3Cos x â€“ 3 = 0

2 â€“ 2CosÂ²x + 3Cos x â€“ 3 = 0

0 = 2CosÂ²x â€“ 3Cos x + 1

0 = 2CosÂ²x â€“ 3Cos x + 1

0 = (2Cos x â€“ 1) (Cos x â€“ 1)

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

Cos x = Â½

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.

( )Â² Â²

(Sec x + 1)Â² = TanÂ²x

SecÂ²x + 2Sec x + 1

= SecÂ²x - 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)Â² = SinÂ² x

Cos x + 1 = Sin x

CosÂ²x + 2Cos x + 1 = 1 â€“ CosÂ² x

2CosÂ² 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 = - Â½

2x =

2x =

x =

x =

Redundant

Answer

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

2Cos 3x - 1 = 0

2Cos 3x = 1

Cos 3x = Â½

3x =

3x =

x =

x =

- Topics covered in this section:
- Solving basic trig equations
- Finding solutions in [0, 2Ï€)
- 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:

SecÂ²x â€“ 3Sec x â€“ 10 = 0

(Sec x + 2) (Sec x â€“ 5) = 0

Sec x + 2 = 0 Sec x â€“ 5 = 0

Sec x = 5

Sec x = -2

Cos x =

Cos x = - Â½

x =

x =

- One of the following equations has solutions and the other two do not. Which equations do not have solutions.
- SinÂ²x â€“ 5Sin x + 6 = 0
- SinÂ²x â€“ 4Sin x + 6 = 0
- SinÂ²x â€“ 5Sin x â€“ 6 = 0
Find conditions involving constants b and c that

will guarantee the equation SinÂ²x + bSin x + c = 0

has at least one solution.

- Find all solutions of the following equation in the interval [0, 2Ï€)
SecÂ²x â€“ 2 Tan x = 4

1 + TanÂ²x â€“ 2Tan x â€“ 4 = 0

TanÂ²x â€“ 2Tan x â€“ 3 = 0

(Tan x + 1) (Tan x â€“ 3) = 0

Tan x = 3

Tan x = -1

Tan x = -1

Tan x = 3

x = ArcTan 3

ref. angle:

71.6Âº

I, III

Quad:

x =

71.6Âº,

251.6Âº