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How to solve for the value of $x$ for the following equation $$\frac {(\sec x + \tan x)^2 - (\sec2x + \tan x)^2}{\sin2x - \sin x} = 2$$

My Attempt: $$ \frac {\sec^{2}x + 2\sec x\tan x + \tan^{2}x - (\frac {1 + \tan^{2}x}{1 - \tan^{2}x} + \tan x)^2}{\frac {2\tan x}{1 + \tan^{2}x} - \sin x} = 2$$

$$ \implies \frac {\sec^{2}x + 2\sec x\tan x + \tan^{2}x - (\frac {1 + \tan^{2}x}{1 - \tan^{2}x})^2 - 2\tan x(\frac {1 + \tan^{2}x}{1 - \tan^{2}x}) - \tan^{2}x}{\frac {2\tan x - \sin x(1 + \tan^{2}x)}{1 + \tan^{2}x}} = 2 $$

But here I stopped. I couldn't find any easier way for simplifying it more as it was growing more complicated while I was trying to solve. My other friends also tried to solve but their final equation turned into complex one.

Clarification: I thought I would lead to a conclusion with bringing an individual trig function. My intention was to solve it via using '$\tan x$' but couldn't transform '$\sec x$' & '$\sin x$' into $'\tan x'$ using any trig identities and formula.

Assumption: There is perhaps a basic in starting point I am missing for holding. Multiplying or dividing with something could have been lot more easier to end with a good conclusion.

Or if that not so, can we move straightforward from the last part of my approach to get the result?

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    $\begingroup$ You can use $\sec x =\pm \sqrt{1+\tan^2x}$ and $\sin x=\pm \frac{\tan x}{\sqrt{1+\tan^2x}}$. $\endgroup$
    – Vishu
    Commented Jan 1, 2021 at 19:26
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    $\begingroup$ Please use \sec and \tan for the trgonometric functions. $\endgroup$ Commented Jan 1, 2021 at 22:19
  • $\begingroup$ @Tavish But applying those formulas are also making the equation bigger. If I use such terms inside root, power of term is getting fractional. If you have another method to solve, I would highly appreciate. $\endgroup$ Commented Jan 2, 2021 at 5:31
  • $\begingroup$ I have nothing else. $\endgroup$
    – Vishu
    Commented Jan 2, 2021 at 8:09
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    $\begingroup$ WolframAlpha is sometimes useful for this sort of thing and shows 9 solutions and a graph here $\endgroup$
    – poetasis
    Commented Jan 2, 2021 at 23:24

1 Answer 1

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Not a pleasant solution but a nasty problem.

Consider that you look for the zero's of function $$f(x)=(\sec (x) + \tan (x))^2 - (\sec (2x) + \tan (x))^2-2(\sin (2x) - \sin(x)) $$ If you try that tangent half-angle substitution, you will end with a polynomial of degree $14$ $$5 t^{14}-3 t^{13}-55 t^{12}-2 t^{11}+193 t^{10}+19 t^9-307 t^8+36 t^7+367 t^6+19 t^5-85 t^4-2 t^3+11 t^2-3 t-1=0$$ which does not factor; so, forget such an approach.

A problem is the discontinuities because of the secants and the tangents. To remove all of them, multiply everything by $\big[\cos^2(x)\, \cos^2(2x)\big]$. Using the multiple angle formula, we end with $$g(x)=9 \sin (x)-2 \sin (2 x)-3 \sin (3 x)-6 \sin (4 x)+5 \sin (5 x)-2 \sin (6x)+$$ $$\sin (7 x)-\sin (8 x)-4 \cos (2 x)+4 \cos (4 x)=0$$ which looks mor pleasant.

In the interval $-\pi \leq x \leq \pi$, there are three obvious solutions $x=-\pi$, $x=0$ and $x=\pi$. Plotting $g(x)$ we can now notice "close" to $x=-0.42$, to $x=0.91$ and to $x=2.15$.

At this point, to polish the roots, I don not see anything elese than Newton method (or any other root-finding procedure). For each root, Newton iterates will be $$\{-0.41,-0.417499,-0.417459\}$$ $$\{0.91,0.907893,0.907886\}$$ $$\{2.15,2.135306,2.135701\}$$

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