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Given 0<=x<2pi, what are the extraneous solutions for the equation cos(x)cot(x)=3-3sin(x)?

3pi/2
0
pi
pi/2

User JosMac
by
8.6k points

1 Answer

3 votes

Answer:

(b) 0 . . . . a possible extraneous solution

(c) pi . . . . a possible extraneous solution

There are no actual extraneous solutions.

Explanation:

You want the extraneous solutions for the equation cos(x)cot(x) = 3 -3sin(x) on the interval [0, 2π).

Extraneous solutions

An extraneous solution is a solution of a transformed equation that is not a solution of the original equation. Extraneous solutions can arise when equations are raised to an even power, or are multiplied by factors that are 0 or ∞ for some values of the variable(s). In general, extraneous solutions are solutions that are excluded from the domain of the original equation.

Exclusions

The given equation is undefined for x an integer multiple of π, that is, for x=0 or x=π. Hence these values are excluded from the domain of the equation. If these show up as a solution, they will be extraneous solutions.

The domain of the equation excludes x = 0 and x = π.

Solution

Multiplying the equation by sin(x), we have ...

cos(x)(cos(x)/sin(x))·sin(x) = (3 -3sin(x))·sin(x) . . . . . multiply by sin(x)

cos(x)² = 3sin(x) -3sin(x)² . . . . . . . . simplify

1 -sin(x)² -3sin(x) +3sin(x)² = 0 . . . . . . . subtract the right-side expression

2sin(x)² -3sin(x) +1 = 0 . . . . . . . . . write in standard form

(2sin(x) -1)(sin(x) -1) = 0 . . . . . . . factor

sin(x) = 1/2 or 1 . . . . . . values of sin(x) that make the product 0

x = π/6, π/2, 5π/6 . . . . . values of x that are solutions

Note that none of these values of x are excluded from the domain, so there are no extraneous solutions.

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Additional comment

Rewriting the equation to the form f(x)=0, the solutions are the x-intercepts of the graph of f(x). These are shown in the attachment. That graph shows the equation is undefined at x=0 and x=π, but those are not potential solutions. None of the solutions we found are extraneous.

Note that this method of solving a equation, finding the zeros of f(x)=0, is usually a good way to avoid extraneous solutions. A graphing calculator cannot show solutions that are not in the domain of f(x).

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Given 0<=x<2pi, what are the extraneous solutions for the equation cos(x)cot-example-1
User Kalehmann
by
8.5k points