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

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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.