Why is 0° not a solution for cosθ = 2sinθ - 1?

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Homework Help Overview

The discussion revolves around the equation cosθ = 2sinθ - 1 and why θ = 0° is not considered a solution. Participants are examining the implications of substituting specific values into the equation and analyzing the results in relation to graphical representations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to substitute θ = 0° into the equation and are questioning the validity of their steps. They are also discussing the discrepancies between calculated solutions and graphical intersections.

Discussion Status

Some participants have provided insights into the reasoning behind why θ = 0° does not satisfy the equation, while others are still seeking clarification on the original poster's confusion regarding their calculations and the graphical results.

Contextual Notes

There are references to potential errors in manipulating the equation, such as squaring both sides, which may introduce extraneous solutions. The discussion also includes specific angles that are proposed as solutions, with some participants expressing uncertainty about their validity.

davon806
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Homework Statement


Please see the attached.
Q1.Why 0° is not a solution of the equation cosθ = 2sinθ - 1?
What's wrong with my steps?
Q2:http://www.wolframalpha.com/input/?i=cosx+=+2sinx+-+1
From the graph,the x-coordinate of the intersection point does NOT equal to the answer
that I calculated in the attached
(0° +360°n or 180° + 360°n or 53.1°+360°n or 126.9° + 360°n),
why?

Homework Equations





The Attempt at a Solution


see the attached.
 

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davon806 said:

Q1.Why 0° is not a solution of the equation cosθ = 2sinθ - 1?
What's wrong with my steps?


Well let's break this up and substitute 0 for θ.

cos0 = 1
2sin0 = 0

So we have 1 = -1, which is clearly not true. I'm really not sure what you were trying to do.
 
I mean when we solve cosθ = 2sinθ - 1,we get
θ = 0° +360°n or 180° + 360°n or 53.1°+360°n or 126.9° + 360°n
That means when θ equals to the above values,LHS(cosθ) = RHS(2sinθ - 1)
But why when we substitute 0 into the above equation,LHS =/= RHS?
And from the graph,the x-coordinate of the intersection point does not equal to the value calculated(0° +360°n or 180° + 360°n or 53.1°+360°n or 126.9° + 360°n),why?
Thx a lot
 
Sorry I have solved Q2 :)
But I have not solved Q1 yet,hope someone can explain to me why x = 0 degree is not the answer.THX :)
 
θ= 0 is not a solution to cos(θ)= 2sin(θ)- 1 because cos(0)= 1 and 2sin(0)- 1= 0- 1= -1, not 1! You apparently squared both sides- any time you do that, or multiply both sides of an equation, you can introduce new solutions that satisfy the new equation but not the original one. To take an obvious example, if you multiply both sides of the equation x= 1 by x, you get [itex]x^2= x[/itex] which now has both x= 1 and x= 0 as solutions.

53.1 degrees is (approximately) a solution. 126.9 is NOT because then the left side of the equation is negative while the right side is positive.
 

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