# Thread: Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

1. ## Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

The full question is:
"Let SIN(α)=a, where α is an element of (0, pi/2). Find, in terms of α, two values of x in [0, 2pi] which satisfies the equation SIN(x)= -a"

I can see that SIN(x)=-SIN(α), but where do I go from here?

We know that α is an angle in the 1st Quadrant, hence SIN(α) is positive.

The answers are: pi+α, and 2pi-α

Thanks in advance for any guidance,
Regards
Smeato

2. ## Re: Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

Have you heard of the CAST rule in trigonometry? You may not have heard it called this. Google it and see.

3. ## Re: Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

Originally Posted by Smeato
The full question is:
"Let SIN(α)=a, where α is an element of (0, pi/2). Find, in terms of α, two values of x in [0, 2pi] which satisfies the equation SIN(x)= -a"
I can see that SIN(x)=-SIN([FONT=arial]α), but where do I go from here?
Surely you know about reference angles. Find the reference angles in $III~\&~IV$ quadrants that are equivalent to the given angle $\alpha$.

4. ## Re: Find solutions for SIN(x)=-a, where a=SIN(α) for 0<α<pi/2

reference angles = $\alpha$ in each quadrant of the unit circle