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

Question

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?

Answer 1

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

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 $n\in \mathbb{Z}$

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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