1. Hi, I've tried to solve this equation:
$(2-\sqrt{2})(1+\cos x)+\tan x=0$
and I've tried everthing but nothing works...Does anybody have an idea?

2. We are given to solve:

$\displaystyle (2-\sqrt{2})(1+\cos(x))+\tan(x)=0$

If we use double-angle identities for cosine and tangent (where $u=\dfrac{x}{2}$), we have:

$\displaystyle (2-\sqrt{2})\left(1+\frac{1-\tan^2(u)}{1+\tan^2(u)}\right)+\frac{2\tan(u)}{1-\tan^2(u)}=0$

$\displaystyle \frac{2-\sqrt{2}}{1+\tan^2(u)}+\frac{\tan(u)}{1-\tan^2(u)}=0$

Multiply through by $1-\tan^4(u)$:

$\displaystyle (2-\sqrt{2})(1-\tan^2(u))+\tan(u)(1+\tan^2(u))=0$

Factor:

$\displaystyle \left(\tan(u)+\sqrt{2}-1\right)\left(\tan^2(u)-\tan(u)+\sqrt{2}\right)=0$

We see that the quadratic factor has complex roots, thus we are left with:

$\displaystyle \tan(u)=1-\sqrt{2}$

Hence:

$\displaystyle u=-\frac{\pi}{8}+k\pi=\frac{\pi}{8}(8k-1)$ where $k\in\mathbb{Z}$

And so:

$\displaystyle x=2u=\frac{\pi}{4}(8k-1)$

3. Originally Posted by MarkFL
If we use double-angle identities for cosine and tangent (where $u=\dfrac{x}{2}$), we have:
Clever!

-Dan

Thanks! There is another solution. When you use double-angle identities for cosine and tangent you have $x\neq \pi+2k\pi$, but this is also a solution of the equation.

5. Originally Posted by laura123
Thanks! There is another solution. When you use double-angle identities for cosine and tangent you have $x\neq \pi+2k\pi$, but this is also a solution of the equation.
Indeed. When switching to the half angle with $u=\frac x2$, we effectively discard the potential solution $x=\pi+2k\pi$, since $\tan(\pm\frac\pi 2)$ is not defined, while $\tan(\pm\pi)$ is defined.
So we need to verify separately if it's a solution... and it is!

Btw, dividing by $1-\tan^2 u$ and multiplying through by $1-\tan^4 u$ are also tricky, since they affect the potential solutions $u=\pm \frac\pi 4 + k\pi$.
However, when verifying them against the original equation, we can see that the solutions found are indeed solutions, and that no other solutions were missed.

6. Originally Posted by laura123
Thanks! There is another solution. When you use double-angle identities for cosine and tangent you have $x\neq \pi+2k\pi$, but this is also a solution of the equation.
Good catch!

I carelessly overlooked any lost solutions.

We can also solve it in this way:

$X=\cos x$
$Y=\sin x$

the equation $(2-\sqrt{2})(1+\cos x)+\tan x=0$ becomes: $(2-\sqrt{2})(1+X)+\dfrac{Y}{X}=0$.
From $\cos^2 x+\sin^2 x=1$ we have $X^2+Y^2=1$, hence:
$$\left\{\begin{array}[l] ((2-\sqrt{2})(1+X)+\dfrac{Y}{X}=0\\X^2+Y^2=1\end{array}\right.$$

thus:

$$\left\{\begin{array}[l] {}X=-1\\Y=0\end{array}\right.\vee \left\{\begin{array}[l] {}X=\dfrac{\sqrt{2}}{2}\\Y=-\dfrac{\sqrt{2}}{2}\end{array}\right.$$
$$\left\{\begin{array}[l] {}\cos x=-1\\\sin x=0\end{array}\right.\Rightarrow x=\pi+2k\pi$$
$$\left\{\begin{array}[l] {}\cos x=\dfrac{\sqrt{2}}{2}\\ \sin x=-\dfrac{\sqrt{2}}{2}\end{array}\right.\Rightarrow x=-\dfrac{\pi}{4}+2k\pi$$