Solving Trig Equations In A Given Range

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Homework Help Overview

The discussion revolves around solving the trigonometric equation cos(3θ) = 0.85 within the specified range of 0 ≤ θ ≤ 360°. Participants are exploring the implications of the cosine function and its behavior in different quadrants.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to solve the equation and has identified several potential solutions based on their understanding of the cosine function and its symmetry. Questions arise regarding the verification of these solutions and the conditions under which cosine is positive.

Discussion Status

Participants are actively engaging with the problem, with some offering insights into the quadrants where cosine is positive. There is an ongoing exploration of the solutions provided by the original poster, and guidance is being shared regarding the general approach to solving the equation.

Contextual Notes

Participants are considering the periodic nature of the cosine function and the implications of the range specified in the problem. There is a mention of potential mistakes in calculations, which highlights the need for careful verification of results.

BOAS
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Homework Statement



Solve cos(3θ) = 0.85 in the range 0 ≤ θ ≤ 360°

Homework Equations





The Attempt at a Solution



cos(3θ) = 0.85

3θ = 31.79°

θ = 10.6° (3 s.f)

I have drawn my transformed cos curve, where the full wave completes after 120°, so there are 3 full cycles in my range.

I often make silly mistakes and I'm not sure of the best method to check my results.

Using the symmetry of the curve I have calculated that θ = 10.6°, 109.4°, 130.6°, 229.4°, 250.6°, 349.4°.

So, how do I verify these?
 
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For what quadrants is the cosine positive?
 
1 and 4
 
Which of your solutions meet this criteria?
 
Hi BOAS! :smile:
BOAS said:
cos(3θ) = 0.85

3θ = 31.79°

Easier if you write:

cos(3θ) = 0.85

3θ = 2nπ ± 31.79° :wink:
 

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