Find $\sin(A)$ and $\cos(A)$ given $\cos^4(A) - \sin^4(A) = \frac{1}{2}$ and $A$ is located in the second quadrant. 
Question: Find $\sin(A)$ and $\cos(A)$, given $\cos^4(A)-\sin^4(A)=\frac{1}{2}$ and $A$ is located in the second quadrant.

Using the fundamental trigonometric identity, I was able to find that:
•　$\cos^2(A) - \sin^2(A) = \frac{1}{2}$
and 
• $$ \cos(A) \cdot \sin(A) = -\frac{1}{4} $$
However, I am unsure about how to find $\sin(A)$ and $\cos(A)$ individually after this.
Edit: I solved the problem through using the Fundamental Trignometric Identity with the difference of $\cos^2(A)$ and $\sin^2(A)$.

Hint
$$\left( \cos(A)+ \sin(A) \right)^2 = 1+2 \sin(A) \cos(A)=\frac{1}{2} \\
\left( \cos(A)- \sin(A) \right)^2 = 1-2 \sin(A) \cos(A)=\frac{3}{2} $$
Take the square roots, and pay attention to the quadrant and the fact that $\cos^4(A) >\sin^4(A)$ to decide is the terms are positive or negative.
Alternate simpler solution
$$2 \cos^2(A)= \left( \cos^2(A)+\sin^2(A)\right)+\left( \cos^2(A)-\sin^2(A)\right)=1+\frac{1}{2} \\
2 \sin^2(A)= \left( \cos^2(A)+\sin^2(A)\right)-\left( \cos^2(A)-\sin^2(A)\right)=1-\frac{1}{2} \\$$
