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

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?

• @Edi No I do not. Apr 9, 2016 at 23:50
• 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. Apr 9, 2016 at 23:50
• Why does it work for this case math.stackexchange.com/a/1698332/298824 Apr 9, 2016 at 23:51
• You shouldn't vandalize your post by erasing the work you did on the problem. Apr 10, 2016 at 0:04
• @N.F.Taussig oops that was an accident Apr 10, 2016 at 0:06

$\sin(4x)−\cos(x)=0$

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

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

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...

• 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 $8\sin^3x - 4\sin x \color{red}{+} 1 = 0$. Apr 10, 2016 at 0:10
• @N.F.Taussig Right, of course, sorry about the typo. Apr 10, 2016 at 0:21