How can I solve the trigonometric equation $(2-\sqrt{2})(1+\cos x)+\tan x=0$?

[laura1231]

Hi, I've tried to solve this equation:

$(2-\sqrt{2})(1+\cos x)+\tan x=0$​

and I've tried everything but nothing works...Does anybody have an idea?

 

[MarkFL]

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:

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

 

[topsquark]

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

Clever! (Bow)

-Dan

 

[laura1231]

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.

 

[I like Serena]

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.

 

[MarkFL]

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! (Yes)

I carelessly overlooked any lost solutions.

 

[laura1231]

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:

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