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

x = 15°

x = 135°

x = -45°

x = -180°

Step-by-step explanation:

Trigonometric Equations

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°

c

step-by-step explanation:

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question 1 answer: p/q can be terminating when q = 5^m * 2^n

question 2 answer: -1 and -1

question 3 answer: -4

question 4 answer: 5/12

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