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?
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?
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)$
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.
Last edited by laura123; November 15th, 2016 at 18:11.
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.
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:
$$
\left\{\begin{array}[l] {}X=-1\\Y=0\end{array}\right.\vee \left\{\begin{array}[l] {}X=\dfrac{\sqrt{2}}{2}\\Y=-\dfrac{\sqrt{2}}{2}\end{array}\right.
$$
$$
\left\{\begin{array}[l] {}\cos x=-1\\\sin x=0\end{array}\right.\Rightarrow x=\pi+2k\pi
$$
$$
\left\{\begin{array}[l] {}\cos x=\dfrac{\sqrt{2}}{2}\\ \sin x=-\dfrac{\sqrt{2}}{2}\end{array}\right.\Rightarrow x=-\dfrac{\pi}{4}+2k\pi
$$