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  1. MHB Apprentice

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    #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?

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    #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)$

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

    -Dan

  4. MHB Apprentice

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    #4 Thread Author
    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.

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    #5
    Quote Originally Posted by laura123 View Post
    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.

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    #6
    Quote Originally Posted by laura123 View Post
    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.

  7. MHB Apprentice

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    #7 Thread Author
    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:

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

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