Solution of the equation below The solution of the equation $\sin 7x + \cos 2x = -2$ is/are?
My Approach: For the above equation to hold true, Both $\sin 7x$ and $\cos 2x$ have to be -1.
$$\sin 7x = -1$$
$$x = \frac{n\pi}{7} + (-1)^n(-\frac{\pi}{14})$$
Also,
$$\cos 2x = -1$$
$$x = n\pi \pm \frac{\pi}{2}$$
But the answer is $2n\pi + \frac{\pi}{2}$
Can anybody help me?
 A: It should be $$7x=-\frac{\pi}{2}+2\pi k$$ and $$2x=\pi+2\pi m,$$ where $\{k,m\}\subset\mathbb Z$, which gives
$$2\left(-\frac{\pi}{2}+2\pi k\right)=7(\pi+2\pi m)$$ or
$$2k-7m=4$$ or
$$2k-6m-4=m.$$
Thus, $m=2n$ and from here
$$2x=\pi+4\pi n$$ or
$$x=\frac{\pi}{2}+2\pi n,$$ where $n\in\mathbb Z$.
A: use that $$\sin(x)+\cos(y)=2 \sin \left(\frac{x}{2}-\frac{y}{2}+\frac{\pi }{4}\right) \sin
   \left(\frac{x}{2}+\frac{y}{2}+\frac{\pi }{4}\right)$$
A: $$7x=2n\pi-\dfrac\pi2=(4n-1)\dfrac\pi2$$
and
$$2x=(2m+1)\pi$$ where $m,n$ are integers
On division,  $$7(2m+1)=4n-1$$
$\iff2(n-2)/7=m$ which is an integer
$\implies7\mid(n-2)$ as $(4,7)=1$
WLOG, $n-2=7r$ where $r$ is any integer
Can you take it from here?
A: As you wrote it, this equation is equivalent to the system of equations
$$\sin 7x =-1,\qquad \cos2x=-1.$$
Now


*

*$\sin 7x=-1 \iff 7x\equiv -\dfrac\pi2\mod 2\pi\iff x\equiv -\dfrac\pi{14}\mod \dfrac{2\pi}7 $,

*$\cos 2x=-1\iff 2x\equiv\pi\mod 2\pi\iff x\equiv\dfrac\pi2\mod\pi$.


This  means we must have integers $k,\ell$ such that
$$x=-\frac\pi{14}+ \frac{2k\pi}7= \frac{(4k-1)\pi}{14}=\frac\pi2+\ell\pi=\frac{(14\ell+7)\pi}{14}.\tag{1}$$
So we have to solve for $4k-1=14\ell+7$ in integers, i.e. $\;2k-7\ell=4$ $(k,\ell \in\mathbf Z)$.
Start from the Bézout's relation $\;2\cdot 4-7=1$. We deduce $\;2\cdot 16-7\cdot 4=4$, so a basic solution is $\;(k,\ell)=(16,4)$, hence the general solution for $k$ and $\ell$:
$$k=16+7u,\quad\ell=4+2u\qquad(u \in\mathbf Z)$$
and finally, replacing in $(1)$: 
$$x=\frac\pi2+\ell\pi=\frac\pi2+(2+u)2\pi,$$
which may be rewritten as $\;\dfrac\pi2+2n\pi$.
