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I was given this problem in my pre-calculus class: "Solve this trig equation for $0 \leq x < \pi$ (All angles are in radians) Equation: $\sin(2x) - \cos(x) = 0$."

Here are the steps I took in my attempt to solve the problem:

  1. I used the trig identity $\sin(2x) = 2\sin(x)\cos(2x)$ to substitute $2\sin(x)\cos(2x)$ for $\sin(2x)$ into the equation to get $2\sin(x)\cos(x) - \cos(x) = 0$.
  2. I added $\cos(x)$ to both sides of the equation to get $2\sin(x)\cos(x) = \cos(x)$, then I divided both sides by $\cos(x)$ to get $2\sin(x) = 1$.
  3. Finally, I divided both sides by $2$ to get $\sin(x) = 1/2$. Since $\arcsin(1/2) = \pi/6$ and $5\pi/6$, I wrote those numbers down as the solutions.

However, my teacher said that there was a third solution, and that I should have factored $2\sin(x)\cos(x) - \cos(x)$ into $\cos(x)[2\sin(x) - 1]$, then set both factors equal to $0$ to get $2\sin(x) - 1 = 0$ and $\cos(x) = 0$. Apparently, the solution I was missing was $\arccos(0) = \pi/2$.

Now after all the background details, here is my question: why didn't my original method find just $2$ of the $3$ solutions? Why couldn't I just get rid of the $\cos(x)$ like I did in step 2?

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  • $\begingroup$ You are not allowed to divide by a quantity that could equal zero. $\endgroup$ Commented Jan 18, 2018 at 19:31
  • $\begingroup$ Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Commented Jan 18, 2018 at 19:37
  • $\begingroup$ $2\sin x\cos x - cos x = 0$ if $\cos x = 0$ so $x = \frac \pi 2$ is a solution. When you divide by $\cos x$ you are assuming $\cos x \ne 0$ so you are ignoring that possible solution. $\endgroup$
    – fleablood
    Commented Jan 18, 2018 at 19:40
  • $\begingroup$ Thanks for all the helpful answers! $\endgroup$
    – Patrick
    Commented Jan 18, 2018 at 20:03
  • $\begingroup$ @PatrickWheeler If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$
    – user
    Commented Jan 22, 2018 at 21:49

4 Answers 4

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we get $$2\sin(x)\cos(x)-\cos(x)=\cos(x)(2\sin(x)-1)=0$$ you can not divide by $\cos(x)$ only when $\cos(x)\neq 0$

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$$\sin 2x-\cos x=0\iff2\sin x \cos x -\cos x=0 \iff \cos x (2\sin x - 1) =0$$ $$\iff \cos x =0 \lor \sin x = \frac12$$

thus

$$x=\frac{\pi}{2}, \quad x=\frac{\pi}{6}, \quad x=\frac{5\pi}{6} $$

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To solve $a_x b_x = a_x c_x$ by dividing both sides by $a_x$ you MUST consider the possibility that $a_x=0$ before you allow yourself to divide both sides.

So you must do:

$a_x b_x = a_x c_x$ then $a_x =0$ will be a possible solution.

If $a_x \ne 0$ then $b_x = c_x$ which will be some more solutions.

So $2\sin x \cos x = \cos x$.

Case 1: If $\cos x = 0$ then $x = \frac \pi 2$ is a solution.

Case 2: If $\cos x \ne 0$ then $2\sin x = 1$ and ......

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Sometimes when one attempts to solve a math problem, one performs operations that change the nature of the problem. This is why one sometimes gets extraneous solutions when solving radical equations. For example, watch what happens when one tries to solve for x when √(x + 4) = x - 2. Soon you will get x² - 5x = 0, which one can use factoring to find x = 0 and x = 5. However, 0 doesn't satisfy the original equation. This is because, somewhere in the process of solving the problem (in this case, when we squared both sides of √(x + 4) = x - 2), we changed the problem enough so that we would get an extraneous solution.

You're original method didn't find all of the solutions because when you removed cos(x) in step 2, you changed the problem enough so that you wouldn't find one of the solutions.

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