Trigonometric problem

Question

I am having trouble solving simple trigonometric equations without a calculator which I am required to be doing in my course since I cant get to understand how to get for example sin x=-1/2 I know sin 30=x so sin -30= -1/2 but what if the domain is only positive how do I get the value the same problem happens to me in the cos and tan functions. Does anyone know something that could help me in that. Thank you.

Answer 1

1. $\sin(2k\pi + x) = \sin((2n+1)\pi - x)$

2. $\cos(2k\pi + x) = \cos(2n\pi - x)$

3. $\tan(2k\pi + x) = \tan((2n+1)\pi + x)$ for all $k$, $n \in \mathbb{Z}$

Using these equations, find the required value.

As an example, consider $\sin(x) = -1/2$.

You know that $x = -30$ satisfies this equation, but since the domain is positive, you know that from the 1st equation for $k=0$, $\sin(x) = \sin((2n+1)\pi - x) = -1/2$, and $x =-30$, the set of solutions is $\{(2n+1)\pi + 30\}$, $n\gt0$

I hope this is clear .