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Ok, while seeing examples in my book concerning with the general solution of trigonometric equations, I saw they used this inequality method . What is this??? The book didn't give any precise elaboration of this method,however it solved many equations with this method. To clarify,some examples:$$ (\tan x)^4 + (\tan y)^4 + 2(\cot x\cot y)^2 = 3 + (\sin(x+y))^2;$$ the book solved this using $$(\tan x)^4 +(\tan y)^4 +2(\cot x\cot y)^2 \geq 4 \text{ and } 3 + {\sin(x+y)}^2 \leq 4$$ in order to have the equality $$(\tan x)^2 = (\tan y)^2 = 1,$$ thus $$x = y = n\pi + \frac{\pi}{4}.$$ Another illustration $$2(\cos x\sin 2x)^2 = x^2 + x^{-2} $$ where
$ 0<x \leq \frac{\pi}{2}$. The inequality method goes like $$2(\cos x\sin 2x)^2 < 2$$ and $$x^2 + x^{-2} \geq 2, $$ so they have no solution.

Still another one, $(\sin x)^6 = 1 + (\cos 3x)^4$, again using inequality method $$(\sin x)^6 \leq 1 \text{ and } 1 + (\cos 3x)^4 \geq 1 , x = (2n + 1)\frac{\pi}{2}.$$ And so many.

But,nothing did the book write regarding this method. When I came before $$\sin 7x = \sin 3x + \sin x \text{, $x$ is in the close interval of 0 and $\pi$},$$ I wanted to use this method,but soon was at the blues as I could not find any way to use this inequality method . My question is what is this inequality method all about? Can I use it in any equation to solve it? If not so,when can I use it? Plz help explaining me what it is & when it can be used .

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  • $\begingroup$ Use backslash to correctly write the trigonometric functions: instead of $\;sin x\;$ , write a backslash immediately before the "s" and get $\;\sin x$ $\endgroup$ – Timbuc Oct 9 '14 at 17:29
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Well, never heard of this method before. It seems to be just some little trick, not really useful in most cases.

The general idea is that you want to solve the equality $f(x)=g(x)$. Then you find such a number $a$ so $\forall x f(x)\le a$ and $\forall x g(x)\ge a$. So, in order to guarantee that the equation holds you need to have $x$ such that $f(x)=g(x)=a$. Now you can solve this equality for $x$. Note that the system $$\begin{cases}f(x)=a,\\g(x)=a,\end{cases}$$ is not always compatible.

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  • $\begingroup$ You seem to be right,sir. But can you tell when I can use this method? Can I solve the last problem with this method? $\endgroup$ – user142971 Oct 10 '14 at 2:06
  • $\begingroup$ I doubt that your last problem can be solved with method. Just try to build the plots for left hand side and right hand side (wolframalpha.com). Anyway, lefthand side is in $[-1,1]$ and the right hand side is non-negative, so such an $a$ as described in my answer does not exist. $\endgroup$ – TZakrevskiy Oct 10 '14 at 12:30

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