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wolverine [178]
3 years ago
10

Solve cos(2x + 30) = 0.5 for -180≤x≤180​

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
6 0

Answer:

<em>x = 15°</em>

<em>x = 135°</em>

<em>x = -45°</em>

<em>x = -180°</em>

Step-by-step explanation:

<u>Trigonometric Equations</u>

Solve

cos(2x + 30°) = 0.5 for -180° ≤ x ≤ 180°

Applying the inverse cosine function:

2x + 30° = arccos(0.5)

There are several angles whose cosine is 0.5. They are 60°, 300°, -60° and -300°, thus we have these candidate solutions:

2x + 30° = 60°

2x + 30° = 300°

2x + 30° = -60°

2x + 30° = -300°

Subtracting 30° to all the equations:

2x = 60° - 30° = 30°

2x = 300° - 30° = 270°

2x = -60° - 30° = -90°

2x = -60° - 300° = -360°

Dividing by 2 we have the complete set of solutions:

x = 15°

x = 135°

x = -45°

x = -180°

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Answer:

See explanation.

Step-by-step explanation:

Statement                                                           Reasons

1) \overline{DC} \cong \overline{BC}1)Given

2) \angle CED \cong \angleD          2)Given

3) \overline{CE} \cong \overline{CD}

and \triangle CED is isosceles      3)Converse of the Base Angle Thm.                      

4) \overline{CE} \cong \overline{CE}4) Reflexive Property

5) \overline{CE} \cong \overline{BC} 5) Transitive Property

6) \angle CBE \cong \angle CEB    

and \triangle CBE is isosceles        6) Base Angle Theorem

I'm going to write my statements and reasons in order below just in case it isn't readable on your computer from above.

Statements:

1) \overline{DC} \cong \overline{BC}

2)  \angle CED \cong \angleD  

3)  \overline{CE} \cong \overline{CD}

and \triangle CED is isosceles  

4) \overline{CE} \cong \overline{CE}

5) \overline{CE} \cong \overline{BC}

6)  \angle CBE \cong \angle CEB    

and \triangle CBE is isosceles

Reasons:

1) Given

2) Given

3) Converse of the Base Angle Theorem

4) Reflexive property

5) Transitive property

6) Base Angle Theorem

---------------------------------------------------------------------------------------------------

Whenever I start a two-column proof, I state me givens.

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Both triangles shared the side CE is why I used the reflexive property.

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I also had that in my given that DC=BC. So by transitive property, I could conclude that CE=BC.  

This is when I finally used the base angle theorem to conclude that the opposite angles of those congruent legs in triangle CBE were congruent.

The base angle theorem says if two legs in a triangle are congruent, then the opposite angles of those legs are congruent.  So you can conclude from this theorem; if is applicable which it was here, that the triangle is an isosceles.

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