Solving cos x + root (1-0.5sin2x) = 0: Check for Spurious Answers

[nokia8650]

"Spurious solutions"

The question involves finding the solution to:

cos x + root (1-0.5sin2x) = 0

I can get the answer; first by taking one term to the other side, squaring, and then factorising.



My answers are:

0,180,45,225

However, the markscheme requir "Checking for spurious answers due to squaring"

Which results in the answers being 180 and 225. Can someone please explain what is meant by this and why, and how one would "check"?

Thanks

 

[dynamicsolo]

When you squared both sides of the equation in order to complete the separation of x, this had the effect of eliminating the relevance of the sign of x. (In a sense, you are no longer solving your original equation.) When you have to do something like this, you need to try your solutions in the original equation to see which ones solve it, rather than your "constructed" equation.

Here's a simple example:

Solve x^2 = x , for real x.

Square both sides:

x^4 = x^2 ,

so (x^2) · [ (x^2) - 1 ] = 0 .

Thus, either x^2 = 0 , giving x = 0 , or

x^2 = 1 , giving x = +/-1 .

Plainly, x = 0 and x = 1 work in the original equation,
but x = -1 does not. This third result solves the "constructed" equation
x^4 = x^2 , but that is not strictly equivalent to the original equation x^2 = x (in fact, we are now really solving for x^2).
So we say that x = -1 is a "spurious solution".

(Of course, had you simply proceeded from (x^2) - x = x·(x - 1) = 0 , you would not have run into this. But this example simply demonstrates the issue -- and was what I came up with off the top of my head. In many situations, you have no option but to square both sides, introducing the difficulty I describe.)

 

[nokia8650]

ahh yes, thank you!

 

[PFStudent]

Hey,

When you squared both sides of the equation in order to complete the separation of x, this had the effect of eliminating the relevance of the sign of x. (In a sense, you are no longer solving your original equation.) When you have to do something like this, you need to try your solutions in the original equation to see which ones solve it, rather than your "constructed" equation.

Here's a simple example:

Solve x^2 = x , for real x.

Square both sides:

x^4 = x^2 ,

so (x^2) · [ (x^2) - 1 ] = 0 .

Thus, either x^2 = 0 , giving x = 0 , or

x^2 = 1 , giving x = +/-1 .

Plainly, x = 0 and x = 1 work in the original equation,
but x = -1 does not. This third result solves the "constructed" equation
x^4 = x^2 , but that is not strictly equivalent to the original equation x^2 = x (in fact, we are now really solving for x^2).
So we say that x = -1 is a "spurious solution".

(Of course, had you simply proceeded from (x^2) - x = x·(x - 1) = 0 , you would not have run into this. But this example simply demonstrates the issue -- and was what I came up with off the top of my head. In many situations, you have no option but to square both sides, introducing the difficulty I describe.)

While I knew this before, I found your explanation very clear and refreshing. Thanks for the post dynamicsolo.

Thanks,

-PFStudent