I think the book is wrong about this trig equation

[srfriggen]

Homework Statement

Solve for x:

Relevant Equations

(sin(2x))/(sec^2(x)) =0

When solving for x I get the angles 0, pi, pi/2 and 3pi/2. However, I thought I should reject the pi/2 and 3pi/2 values since they are not in the domain of sec^2(x). I am using the opens tax precalc book and their answer does not reject those two angles.

 

[anuttarasammyak]

observing\cos x = \frac{1}{\sec x}your concern may disappear.

 

[Mark44]

Homework Statement:: Solve for x:
Relevant Equations:: (sin(2x))/(sec^2(x)) =0

When solving for x I get the angles 0, pi, pi/2 and 3pi/2. However, I thought I should reject the pi/2 and 3pi/2 values since they are not in the domain of sec^2(x). I am using the opens tax precalc book and their answer does not reject those two angles.

I get your concern and agree that those values should be rejected. For $x = \frac \pi 2$ or $x = \frac {3\pi} 2$, the denominator in your equation is zero, so the fraction is undefined.

observing $\cos ⁡x=\frac 1 {\sec⁡ x}$ your concern may disappear.

Perhaps this is what the authors of the solution did, but nevertheless, the two values I mentioned make the original equation undefined on the left side.

 

[srfriggen]

observing\cos x = \frac{1}{\sec x}your concern may disappear.

The denominator of the original fraction becomes zero at the two values I mentioned. I used this fact to solve the equation but then rejected those two values. Should we not take into account the order of operations of the input angle in the original composite function, not the manipulated equation?

 

[anuttarasammyak]

It is too subtle for me to distinguish
\sin 2x \cos^2x=0 and
\sin 2x / \sec^2x=0
or
x-1=0 and
\frac{1}{\frac{1}{x-1}}=0
and hope that it does not matter for physics.

 

[srfriggen]

The first two equations you wrote do not have the same domains. Consider y=(sqrt(x))^2 and y=x. The former has a domain [0, inf) white the latter is all Reals, even though the equations are both equal.

 

[srfriggen]

I get your concern and agree that those values should be rejected. For $x = \frac \pi 2$ or $x = \frac {3\pi} 2$, the denominator in your equation is zero, so the fraction is undefined.
Perhaps this is what the authors of the solution did, but nevertheless, the two values I mentioned make the original equation undefined on the left side.

Thank you for the reply

 

[anuttarasammyak]

Consider y=(sqrt(x))^2 and y=x. The former has a domain [0, inf) white the latter is all Reals, even though the equations are both equal.

x=sgn(x)|x|=sgn(x)\sqrt{x^2}\neq \sqrt{x^2}
This confirms that the equations
y-x=0
and
y-\sqrt{x^2}=0
are not equal.

Many texts e.g. Wikipedia article of trigonometry says
\sec x = \frac{1}{\cos x}
You should add "except x=half integer pi" in mind for all those statements. It seems tough work for me, a lazy guy. I am tempted to write
\frac{1}{\sec{\pi/2}}=0
or more directly
\frac{1}{\pm \infty}=0

 

[PeroK]

Homework Statement:: Solve for x:
Relevant Equations:: (sin(2x))/(sec^2(x)) =0

When solving for x I get the angles 0, pi, pi/2 and 3pi/2. However, I thought I should reject the pi/2 and 3pi/2 values since they are not in the domain of sec^2(x). I am using the opens tax precalc book and their answer does not reject those two angles.

Technically, of course, you are correct. Mathematically, in general we have:$$\frac{f(x)}{g(x)} = 0 \ \Leftrightarrow \ f(x) = 0$$That said, if we define the funtions: $$t_1(x) = \frac{\sin(2x)}{\sec^2 x} \ \text{and} \ t_2(x) = \sin(2x)\cos^2 x$$then we see that $t_1(x) = t_2(x)$ except where $\cos x = 0$, where $t_1(x)$ is not defined. And we see that the discontinuities in $t_1(x)$ are all removable. We can, therefore, view $t_2(x)$ as a patched-up version of $t_1(x)$ with the removable discontinuities removed.

In that sense, we can recast the question in terms of $t_2(x)$ and obtain the additional solutions. In many cases, this would be a mathemetically important thing to do. Especially if we question how we ended up with $\sec^2 x$ in the denominator instead of $\cos^2 x$ in the numerator in the first place!

 

[Mark44]

x=sgn(x)|x|=sgn(x)\sqrt{x^2}\neq \sqrt{x^2}
This confirms that the equations
y-x=0
and
y-\sqrt{x^2}=0
are not equal.

Equations are not "equal" to each other. The proper term is "equivalent," meaning that two equivalent equations have exactly the same solution sets.

Just as your example shows that the equations $y = x$ and $y = \sqrt{x^2}$ are not equivalent, it's also true that $y = \frac{\sin(2x)}{\sec^2(x)}$ and $y = \sin(2x) \cos^2(x)$ are not equivalent, the equations in post #1 of this thread.

Many texts e.g. Wikipedia article of trigonometry says
\sec x = \frac{1}{\cos x}
You should add "except x=half integer pi" in mind for all those statements. It seems tough work for me, a lazy guy. I am tempted to write
\frac{1}{\sec{\pi/2}}=0
or more directly
\frac{1}{\pm \infty}=0

It's not legitimate to use infinity in any kind of arithmetic or algebraic expression.