7
$\begingroup$

Solve $\cos(7x)+\sin(3x)=0$

I'm stuck. Help me, please.

I did $\cos(7x)=\cos(4x+3x)=\cos(4x)\cos(3x)-\sin(4x)\sin(3x)$

So the original equation becomes

$\cos(4x)\cos(3x)-\sin(4x)\sin(3x)+\sin(3x)=0$

But that became long and ugly. What can I do now?

$\endgroup$
12
$\begingroup$

$$\displaystyle \cos (7x) = -\sin (3x) = \sin (-3x) = \cos \left(\frac{\pi}{2}+3x\right)$$

So we get $$\displaystyle \cos (7x) = \cos \left(\frac{\pi}{2}+3x\right)$$

Using $\displaystyle \cos x= \cos \alpha\;,$ Then $\displaystyle x=2n\pi\pm \alpha\;,$ Where $n\in \mathbb{Z}$

So We get $$\displaystyle 7x=2n\pi\pm \left(\frac{\pi}{2}+3x\right)\;,$$ Where $n\in \mathbb{Z}$

$\endgroup$
  • $\begingroup$ I believe you mean $n$ instead of $x$ in the final line $\endgroup$ – Brenton Oct 2 '15 at 18:58
  • $\begingroup$ Sorry Brenton I have edited it. $\endgroup$ – juantheron Oct 2 '15 at 18:59
  • $\begingroup$ No need to apologize. It's a great solution! +1 $\endgroup$ – Brenton Oct 2 '15 at 19:00
3
$\begingroup$

You have $$\cos (7x)=-\sin (3x)=\sin(-3x)=\cos (90--3x)$$ Hence, $$7x=\pm(90+3x)+n.360, n\epsilon\mathbb{Z}$$

$\endgroup$
3
$\begingroup$

Well, since no one else has, here's the prosthaphaeresis solution: we have $$ \cos{7x}+\cos{(\tfrac{1}{2}\pi-3x)} = 0, $$ using $\cos{x}=\sin{(\frac{1}{2}\pi-x)}$. We then have the prosthaphaeresis formula $$ \cos{A}+\cos{B} = 2\cos{\frac{A+B}{2}}\cos{\frac{A-B}{2}}, $$ which we use to write the equation in the form $$ 0 = \cos{\left( \frac{7x-3x+\pi/2}{2} \right)}\cos{\left( \frac{7x+3x-\pi/2}{2} \right)} = \cos{\left( 2x+\frac{\pi}{4} \right)}\cos{\left( 5x-\frac{\pi}{4} \right)} $$ Therefore the equation is satisfied precisely when one of these factors is zero. $\cos{A}$ has roots when $A=(n+1/2)\pi$ for integer $n$, and I'm sure you can take it from here.

$\endgroup$
  • $\begingroup$ Prosthaphaeresis? Whoah! I just call them sum-to-prduct formula $\endgroup$ – G-man Oct 2 '15 at 19:29
  • $\begingroup$ Then I'm happy I taught you a new word. Rolls right off the tongue, no? Wikipedia has some information on their historical uses. $\endgroup$ – Chappers Oct 2 '15 at 19:31
2
$\begingroup$

Notice that $\cos 7x+\sin 3x=\cos(5x+2x)+\sin(5x-2x)$.

Expand the sine and cosine and then gather terms to get this equation: $$(\sin5x+\cos5x)(\cos2x-\sin2x)=0$$ Now either $\tan5x=-1$ or $\tan2x=1$.

The first case leads to the solution $x=\frac{\pi}{20}\cdot(4n-1)$ and the second leads to $x=\frac{\pi}8(4m+1)$.

The solution to your equation would be the union of the two.

$\endgroup$
0
$\begingroup$

HINT: Show that $$2 \sin \left(\frac{\pi }{4}-2 x\right) \sin \left(x+\frac{\pi }{4}\right) (2 \sin (2 x)+2 \cos (4 x)-1)=0$$

$\endgroup$

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.