TITLE: Solving $\sin\left(2x\right)+5\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)+1=0$
QUESTION [8 upvotes]: I attempted to do this trig problem I have written below with steps I took to reach an answer. Did I do this correctly?
$$\sin\left(2x\right)+5\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)+1=0$$
using the well known identity: $\sin^2\left(x\right)+\cos^2\left(x\right)=1$, I get:
$$\color{red}{\sin\left(2x\right)}+5\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)+\color{red}{\sin^2\left(x\right)+\cos^2\left(x\right)}=0$$
Then I see that the "red" is just $\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)^2$, so
$$\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)^2+5\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)=0$$
Then I let $u=\sin\left(x\right)+\cos\left(x\right)$, I get
$$u^2+5u=0$$
$$u=0,\:\text{and}\:u=-5$$
removing $u=-5$ as no solution, leaving just $u=0$ as the only solution, plugging back into the defined "$u$",
$$0=\sin\left(x\right)+\cos\left(x\right)$$
$$\sin\left(x\right)=-\cos\left(x\right)$$
$$\tan\left(x\right)=-1$$
the solution I get is
$$x=\frac{3\pi }{4}+\pi n$$
No need to see if there are any extraneous solutions since the function is continuous.
Thanks for any feedback!

REPLY [3 votes]: Overall, looks great, though I've a few things to comment on.

My first criticism is that you should perhaps justify the following:

$(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + \sin 2x$. This is easily done by squaring the binomial by noting $\sin 2x = 2 \sin x \cos x$.

$u = -5$ generates no solution. This follows easily since $\sin x, \cos x$ are, at minimum, $-1$ each.


Granted, as noted, these are simple justifications (clearly you already know them), but perhaps would be appreciated by your instructor for completeness (if this is for a class).

Next, towards the end, you divide by $\cos x$ in order to get $\tan x$. This results in a problem if $\cos x = 0$, though. Thus, you need to handle that case separately, and need to determine whether or not $x$ can be such that both sine and cosine are equal to zero.
(Obviously, they're not, and you have the complete solution set -- this is just, again, in the interest of completeness.)