# General solutions of a trigonometric equations

Consider Sin(-A)= 0.5

I want to find the general solution for A.

Method 1

Negativity is relative $\{5=-(-5),\}$ So, I could write -A as B, so I can rewrite my question as

Sin(B)= 0.5

This looks pretty straightforward, hence I write the general solution for B,

$$B=nπ+(-1)^n(π/6)$$ so,$$-A=nπ+(-1)^n(π/6)$$ $$A=-\{nπ+(-1)^n(π/6)\}$$

Method 2

Sin(-A) = 0.5

-Sin(A)=0.5

Sin(A) =-0.5

Now I write the general solution, $$A=nπ+(-1)^n(-π/6)$$

As Sin(-A)= Sin(A) These two solutions must refer to the same A, So, I can equalize the above general solutions

Then if I solve the equation for n, I get that the equality holds only when n=0, but these are general solutions, how can this be?

I am very confused. Pls help me out, Thanks

The function sinus is odd i. e. $\sin (-x)=-\sin(x)$ as you know. So, since $\arcsin \left(\dfrac 12\right)=\dfrac{\pi}{6}$ all you have to do is $$A=-\dfrac{\pi}{6}+2n\pi$$