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

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In summary: From the graph,the x-coordinate of the intersection point does NOT equal to the answer (0° +360°n or 180° + 360°n or 53.1°+360°n or 126.9° + 360°n),why?The x-coordinate of the intersection point does not equal the answer because when θ equals to the above values,LHS(cosθ) = RHS(2sinθ - 1) and when θ equals to 0°,LHS= RHS.
  • #1
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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  • #2
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.
 
  • #3
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
 
  • #4
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 :)
 
  • #5
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.
 
  • #6

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

1. What are trigonometric ratios?

Trigonometric ratios are mathematical functions that relate the sides and angles of a right triangle. The three main trigonometric ratios are sine, cosine, and tangent.

2. How do you solve trigonometric ratio problems?

To solve a trigonometric ratio problem, you must first identify the given information and the unknown value. Then, use the appropriate trigonometric ratio (sine, cosine, or tangent) to set up an equation and solve for the unknown value using algebraic methods.

3. What is the Pythagorean theorem?

The Pythagorean theorem is a mathematical equation that states the relationship between the sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

4. How do trigonometric ratios relate to real-world problems?

Trigonometric ratios are used in various fields such as engineering, physics, and navigation to solve real-world problems involving angles and distances. For example, they can be used to calculate the height of a building or the distance between two objects.

5. What are some common mistakes to avoid when solving trigonometric ratio problems?

Some common mistakes to avoid when solving trigonometric ratio problems include using the wrong trigonometric ratio, not converting angles to the correct unit, and not using proper algebraic steps. It is also important to check your solutions and ensure they make sense in the context of the problem.

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