# Thread: Trigonometric equations with parameter

1. ## Trigonometric equations with parameter

How would you solve this equation, and most importantly how would you have written down the answer to this equation?

$\sin 2x + \sin x = a + \cos x$, where $a$ is a parameter.

I solved this equation correct, I think, but how to write the final answer, I can not understand.

2. Originally Posted by DeMath
How would you solve this equation, and most importantly how would you have written down the answer to this equation?

$\sin 2x + \sin x = a + \cos x$, where $a$ is a parameter.

I solved this equation correct, I think, but how to write the final answer, I can not understand.
the only important point is that we must have $-1-\sqrt{2} \leq a \leq \frac{5}{4}.$ here's why: let $\sin x - \cos x = t.$ then you get $1+t-t^2=a,$ i.e. $5-4a=(2t-1)^2.$

we also know that $-\sqrt{2} \leq t \leq \sqrt{2},$ which proves the claim. now you can write $\sin x - \cos x = t=\frac{1 \pm \sqrt{5-4a}}{2},$ which gives you $\sin \left(x - \frac{\pi}{4} \right) = \frac{1 \pm \sqrt{5-4a}}{\sqrt{2}}.$

therefore $x =\frac{\pi(4n+1)}{4} + (-1)^n \sin^{-1} \left( \frac{1 \pm \sqrt{5-4a}}{\sqrt{2}} \right), \ \ n \in \mathbb{Z}.$

3. Originally Posted by NonCommAlg
the only important point is that we must have $-1-\sqrt{2} \leq a \leq \frac{5}{4}.$ here's why: let $\sin x - \cos x = t.$ then you get $1+t-t^2=a,$ i.e. $5-4a=(2t-1)^2.$

we also know that $-\sqrt{2} \leq t \leq \sqrt{2},$ which proves the claim. now you can write $\sin x - \cos x = t=\frac{1 \pm \sqrt{5-4a}}{2},$ which gives you $\sin \left(x - \frac{\pi}{4} \right) = \frac{1 \pm \sqrt{5-4a}}{\sqrt{2}}.$

therefore $x =\frac{\pi(4n+1)}{4} + (-1)^n \sin^{-1} \left( \frac{1 \pm \sqrt{5-4a}}{\sqrt{2}} \right), \ \ n \in \mathbb{Z}.$
This equalization was offered on preliminary examinations in one of the Moscow universities, but nobody got a higher estimation for this equalization, although many decided it exemplary as well as you.

This equation is proposed for the entrance exams at one of Moscow universities. But no one got the highest mark for this equation, although many have decided it is just like you. In an official statement after the exams were told that applicants applying for appreciation were to consider a few cases, and based on this wrote roots, which are not duplicated.

I think the answer should be written in such a way

$\text{If}~~a \in\!\Bigl[-1-\sqrt 2 ;\sqrt 2 -1\Bigl)~~ \text{then}~~ x = (-1)^k \arcsin \frac{\sqrt 2 -\sqrt {10-8a}}{4} + \frac{\pi}{4}(4k+1),~k \in \mathbb{Z}.$

$\text{If}~~a \in\!\left[\sqrt 2 -1;\,\frac{5}{4}\right]~~ \text{then}~~x = (-1)^k \arcsin \frac{\sqrt 2 \pm\sqrt {10-8a}}{4} + \frac{\pi}{4}(4k+1),~k \in \mathbb{Z}.$

But one professor said that we should consider more cases.