Solving for x Algebraically So my problem is:
$$\arcsin (x) = \arccos (5/13)$$ 
^ Solve for $x$.
How would I begin this problem? Do I draw a triangle and find the $\sin(x)$ or is there a more algebraic way of doing this? Thanks in advance for any help.
 A: Let $\arccos\frac5{13}=\arcsin x=\theta\implies x=\sin\theta,\cos\theta=\frac5{13}$
Using this, $0<\theta<\dfrac\pi2\implies\sin\theta>0$
Use $\sin^2\theta+\cos^2\theta=1$
A: You do draw a triangle, but "$x$" does not represent something on that triangle that you would take the sin of. Think instead of what $\arcsin(x)$ represents: it is an angle of the triangle. So, assign an angle variable $\theta = \arcsin(x)$. Your job, if you decide to accept it, is to evaluate the angle $\theta$. 
Draw a right triangle. Assign one of its angles (not the right angle) to be $\theta$. Assign letters for its side lengths. Write out the equation for the Pythagorean formula of a right triangle using your letters for side lengths.
You are told that $\theta = \arcsin(x)$, so $\sin(\theta)=x$. Hence the ratio $opposite/hypotenuse$ equals $x$. Write out that equation using your letters for side lengths. 
You are also told that $\theta = \arccos(5/13)$, so $\cos(\theta) = 5/13$, hence the ratio $adjacent / hypotenuse$ equals $5/13$. Write out that equation using your letters for side lengths. 
Now you have a bunch of equations, and you should be able to eliminate all the other variables (the side lengths) and solve the equations for $\theta$.
