When solving $\sin(2\theta) = \cos\theta$ for $\theta$ for all values in the range $[0,2\pi]$ I only get half of the solutions when I reduce $\sin(2\theta) = \cos\theta$ to $\sin\theta = \frac{1}{2}$.

So before reducing the equation the solutions are $\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$ and after reducing the equation the solutions are only $\frac{\pi}{6}, \frac{5\pi}{6}$

I can see how it reduces the solution set, but I want to know 'why' it happens.

How can you know exactly when you're reducing the solution set of an equation when you're reducing the equation itself to a simpler form, especially when it's a big complicated equation when you can't easily see that you're reducing the solution set as in this case?

Is there some information regarding that or am I asking the wrong question?

  • 4
    $\begingroup$ You cancelled a factor of an unknown value. When that factor happens to be zero, the original equation holds, but what you were left with may not... $\endgroup$ – Jyrki Lahtonen Jun 17 '14 at 20:27
  • $\begingroup$ I feel confident about the fact that removing the possibility of a 'zero solution' takes away the possibility of that being part of the solution set, but somehow I still don't feel very satisfied with that answer. $\endgroup$ – Kermit the Hermit Jun 17 '14 at 20:35
  • $\begingroup$ Maybe it's because I am so used to school always giving me equations that retain their solution sets even after reducing them. $\endgroup$ – Kermit the Hermit Jun 17 '14 at 20:36
  • $\begingroup$ Would it always only be the 'zero solution' that stands the chance of being removed after reducing the equation or is there more to it? $\endgroup$ – Kermit the Hermit Jun 17 '14 at 20:38

$$\sin(2\theta) = \cos \iff 2\sin\theta \cos\theta - \cos \theta = 0 \iff \cos \theta(2 \sin\theta - 1) = 0$$ $$\iff \cos \theta = 0 \text{ or } \sin\theta = \frac 12$$

If you canceled $\cos \theta$ from each side of the equation (which is legitimate only for $\cos \theta \neq 0$), then you need to additionally consider the solution to when $\cos \theta = 0$, otherwise you lose that information by canceling that factor.

  • $\begingroup$ Okay okay this comforts me. So what advice would you give me to prevent losing such information in the future, factorize? $\endgroup$ – Kermit the Hermit Jun 17 '14 at 21:00
  • 1
    $\begingroup$ You can cancel a factor, provided it doesn't equal zero. Remember, when you cancel a factor on each side of an equation, you are dividing each side of that equation, which you can't do if, in this case, $\cos \theta = 0$. Whenever you're tempted to cancel a factor on each side of an equation, make a note, in your work, to ensure you check when that factor is $0$. If you did that here, you'd have seen that indeed, $\cos x = 0$ is a solution: You get $0 = 0$, which is true. Otherwise, aim to move everything to the left, so that the right-hand side = 0. $\endgroup$ – Namaste Jun 17 '14 at 21:03
  • $\begingroup$ You've totally cleared this up for me. Thanks a lot. $\endgroup$ – Kermit the Hermit Jun 17 '14 at 21:12
  • $\begingroup$ You're welcome, Midni. $\endgroup$ – Namaste Jun 17 '14 at 21:27

$$\sin(2\theta)=\cos \theta\iff2\sin\theta\cos\theta=\cos\theta\iff(2\sin\theta=1)\lor\boxed{(\cos\theta=0)}$$

  • $\begingroup$ This is almost what you posted but on short way which helps to memorize it:-)@amWhy $\endgroup$ – user63181 Jun 17 '14 at 20:35
  • $\begingroup$ Is that always the case when you reduce an equation that you maintain the solution when the variable equals zero and then add it to the solution set of the reduced equations solution set? $\endgroup$ – Kermit the Hermit Jun 17 '14 at 20:48
  • $\begingroup$ If we have the equality $ab=ac$ then we write it $ab-ac=a(b-c)=0$ so $a=0$ or $b-c=0$ hence $a=0$ or $b=c$. $\endgroup$ – user63181 Jun 17 '14 at 20:51

If $x=\cos\theta$ and $y=\sin\theta$ then $\sin2\theta = 2xy$ and your equation is:

$$2xy = x$$ or $$(2y-1)x=0$$ That's true when $y=\frac{1}{2}$ and when else?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.