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

Question

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?

Am I correct or not?

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. — colormegone, Apr 09 '16 at 23:50
Why does it work for this case http://math.stackexchange.com/a/1698332/298824 — bigfocalchord, Apr 09 '16 at 23:51
You shouldn't vandalize your post by erasing the work you did on the problem. — N. F. Taussig, Apr 10 '16 at 00:04
Your solution is correct. If you click on more roots on Wolfram alpha, you will find the solutions $\frac{\pi}{6} + 2n\pi$ and $\frac{5\pi}{6} + 2n\pi$. Together with the solution $n\pi - \frac{\pi}{2}$, those solutions are the solutions to the equation $x = \frac{\pi}{6} + \frac{2n\pi}{3}$. The other solutions listed above together with $x = n\pi - \frac{\pi}{2}$ are equivalent to your solution $x = \frac{\pi}{5} + \frac{2n\pi}{5}$. See this explanation. — N. F. Taussig, Apr 10 '16 at 00:40
Everything is fine. Your solution is quite nice and compact. — André Nicolas, Apr 10 '16 at 00:43
@AndréNicolas thank you for the confirmation, is there a trick to know if solutions overlap or not? — bigfocalchord, Apr 10 '16 at 00:48
At one point I was writing up an answer to the effect they are the same. — André Nicolas, Apr 10 '16 at 01:43

KR136 · Accepted Answer · 2016-04-10T00:20:40.150

3

$\sin(4x)−\cos(x)=0$

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

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

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...

edited Apr 10 '16 at 00:20

answered Apr 09 '16 at 23:52

KR136

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 $8\sin^3x - 4\sin x \color{red}{+} 1 = 0$. – N. F. Taussig Apr 10 '16 at 00:10
@N.F.Taussig Right, of course, sorry about the typo. – KR136 Apr 10 '16 at 00:21

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