# Trigonometric equation with two functions and a parameter

The problem: For which values of $a$ the equation $a.sinx.cosx=sinx-cosx$ has exactly $2$ different roots in the interval $[0;\pi]$

So clearly $0,\frac{\pi}{2},\pi$ are not an answer. I divide by $sinx.cosx$ and get $f(x)=\frac{1}{cosx}-\frac{1}{sinx}=a$. I look at the interval $(0;\frac{\pi}{2})$ $cosx$ is decreasing so $\frac{1}{cosx}$ is increasing same for $sinx$ is increasing and $\frac{-1}{sinx}$ is also increasing. $\displaystyle \lim_{x \to 0}f(x)= +\infty$ and $\displaystyle \lim_{x \to\frac{\pi}{2} }f(x)= -\infty$ and $f(x)$ is continuous so we have exactly $1$ root in the interval $(0;\frac{\pi}{2})$ for every $a$.

My question: I guess I have to find for what value of $a$ the function has $1$ root in the interval $(\frac{\pi}{2};\pi)$ how do I do that and is there an easier way to solve the whole problem?

It may be useful to look at the graphs of the given functions. The given equation can be converted to the following, $$a\sin{2x}=2\sqrt{2}\sin{(x-\frac{\pi}{4})}$$ Now sketch the graphs and observe that you will have two solutions iff $a=-2\sqrt{2}$. Here one solution will lie in $(0,\pi/2)$ where the graphs will intersect transversely, whereas one more solution in $(\pi/2,\pi)$ where the graphs will 'touch' each other tangentially. Now you can play with the values of $a$ and see that the number of solutions will increase and decrease if you change $a$ slightly.
Let $f(x)=\frac{1}{cosx}-\frac{1}{sinx}, x\in (0, \frac {\pi}2)$. Then $f$ is strictly increasing and $\displaystyle \lim_{x \to 0}f(x)= -\infty, \displaystyle \lim_{x \to \frac {\pi} 2}f(x)= +\infty$ therefore the equation $f(x)=a$ has a unique solution on $(0, \frac {\pi}2)$
For $x \in (\frac {\pi}2, \pi)$, then $\displaystyle \lim_{x \to {\frac {\pi} 2}}f(x)= -\infty$ and $\displaystyle \lim_{x \to {\pi}}f(x)= -\infty$. To have a unique solution in $(\frac {\pi}2, \pi)$, we must take $a$ the maximum value of $f$. Can you take it from here?
• I dont understand tanking $a$ the max of $f$ – yolo expectz Mar 27 '17 at 19:58
• @yoloexpectz If $a$ is not the max, you'll have at least two values of $x$ such that $f(x) = a$. The line $y=a$ must be tangent to the graph of $f$ – user261263 Mar 27 '17 at 20:01