1
$\begingroup$

Question: Find the general solution to $\sin(4x)-\cos(x)=0$


My attempt:

$$ \sin(4x)-\cos(x)=0$$

$$ \Leftrightarrow \sin(4x) = \cos(x) $$

$$ \Leftrightarrow \sin(4x) = \sin( \frac{\pi}{2} -x)$$

From another post I learnt that you can equate $\sin(x) = \sin(y)$ on 2 conditions so applying it here:

$$ \Leftrightarrow 4x = \frac{\pi}{2} -x + 2 \pi n$$

and $$ \Leftrightarrow4x = \pi - (\frac{\pi}{2} -x)+ 2 \pi n$$

Solving both for $x$

$$ x = \frac{\pi}{10} + \frac{2\pi n }{5}$$

$$ x = \frac{\pi}{6} + \frac{2\pi n }{3}$$

However I checked Wolfram alpha and they have different solutions?

enter image description here

Am I correct or not?

$\endgroup$

closed as off-topic by bigfocalchord, colormegone, Shailesh, Claude Leibovici, user91500 Apr 10 '16 at 7:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – colormegone, Shailesh, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ @Edi No I do not. $\endgroup$ – bigfocalchord Apr 9 '16 at 23:50
  • $\begingroup$ You can't directly compare trig functions of different periods this way: you will have to find an expression for $ \ \sin 4x \ $ in terms of $ \ \sin x \ $ and $ \ \cos x \ $ using the double-angle formula for sine twice and other trig identities as well. $\endgroup$ – colormegone Apr 9 '16 at 23:50
  • $\begingroup$ Why does it work for this case math.stackexchange.com/a/1698332/298824 $\endgroup$ – bigfocalchord Apr 9 '16 at 23:51
  • $\begingroup$ You shouldn't vandalize your post by erasing the work you did on the problem. $\endgroup$ – N. F. Taussig Apr 10 '16 at 0:04
  • $\begingroup$ @N.F.Taussig oops that was an accident $\endgroup$ – bigfocalchord Apr 10 '16 at 0:06
3
$\begingroup$

$\sin(4x)−\cos(x)=0$

$2\sin(2x)\cos(2x)-\cos(x)=0$

$4\sin(x)\cos(x)(1-2\sin^2(x))-\cos(x)=0$

One possible solution is $\cos(x)=0$

$4\sin(x)(1-2\sin^2(x))=1$

$8\sin^3(x)-4\sin(x)+1=0$

Now, let $\sin(x)=m$ and solve the resulting cubic...

$\endgroup$
  • $\begingroup$ Observe that $$4\sin x\cos x(1 - 2\sin^2x) - \cos x = \cos x(4\sin x - 8\sin^3x - 1)$$ so we obtain $\cos x = 0$ or $4\sin x - 8\sin^3x - 1 = 0$. If we multiply the cubic polynomial in sine by $-1$, we obtain $8\sin^3x - 4\sin x \color{red}{+} 1 = 0$. $\endgroup$ – N. F. Taussig Apr 10 '16 at 0:10
  • $\begingroup$ @N.F.Taussig Right, of course, sorry about the typo. $\endgroup$ – KR136 Apr 10 '16 at 0:21

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