Trigonometric inequality help

[dolina dahani]

Hello there folks, I am new since I really have to get an answer. So I am not sure if I'm stupid or what. The question is:
Solve trigonometric inequality in the real numbers ( I mean set of real numbers)
sin(x)+√3cos(x)>0
Sorry the book is in Serbian so please don't mind that, it goes with addiutional formulas, but I don't understand how to get that multiplication with 1/2, as I know we can add but never multiply in inequalities. I understand everything after that just that first but is annoying. And is there any other way to solve this. Thanks!
THE QUESTION IS UNDER d)

 

[MarkFL]

Hello, and welcome to FMH!

They have essentially used a linear combination identity:

[MATH]\sin(x)+\sqrt{3}\cos(x)=\sqrt{1^2+(\sqrt{3})^2}\sin\left(x+\arctan\left(\frac{\sqrt{3}}{1}\right)\right)=2\sin\left(x+\frac{\pi}{3}\right)[/MATH]
And then dividing by 2, we get:

[MATH]\sin\left(x+\frac{\pi}{3}\right)>0[/MATH]
This is how I would approach the problem. And then with \(k\in\mathbb{Z}\), I would write:

[MATH]2k\pi<x+\frac{\pi}{3}<(2k+1)\pi[/MATH]
[MATH]2k\pi-\frac{\pi}{3}<x<2k\pi+\frac{2\pi}{3}[/MATH]

 

[dolina dahani]

---

Hello, and welcome to FMH!

They have essentially used a linear combination identity:

[MATH]\sin(x)+\sqrt{3}\cos(x)=\sqrt{1^2+(\sqrt{3})^2}\sin\left(x+\arctan\left(\frac{\sqrt{3}}{1}\right)\right)=2\sin\left(x+\frac{\pi}{3}\right)[/MATH]
And then dividing by 2, we get:

[MATH]\sin\left(x+\frac{\pi}{3}\right)>0[/MATH]
This is how I would approach the problem. And then with \(k\in\mathbb{Z}\), I would write:

[MATH]2k\pi<x+\frac{\pi}{3}<(2k+1)\pi[/MATH]
[MATH]2k\pi-\frac{\pi}{3}<x<2k\pi+\frac{2\pi}{3}[/MATH]

Thank you sir, for the help