When dealing with inverses, it is best to draw an accompanying triangle.
cos^-1(t) = sin^-1(t)
cos(cos^-1(t)) = cos(sin^-1(t))
t = cos(sin^-1(t))
Now focus on sin^-1(t), this means that sin(θ) = t. Draw a right triangle. Label one of the angles θ. The opposite side of that angle is t and the hypotenuse is 1. Then by the pythagorean theorem, you get the adjacent side to be √(1 - t^2). Since you drew a triangle and sin^-1(t) refers to θ in that triangle, then cos(sin^-1(t)) = cos(θ) = √(1 - t^2). Now you get:
t = √(1 - t^2)
Square both sides to get:
t^2 = 1 - t^2
2t^2 = 1
t^2 = 1/2
t = ±1/√2
Because funny things can happen when when you square square roots (it has to do with range/domain restrictions on functions) you need to check your answers for extraneous solutions. You'll see that the only solution that works is 1/√2, that is the correct answer.
2tan^-1(x) = tan^-1(x/4)
tan(2tan^-1(x)) = tan(tan^-1(x/4))
x/4 = tan(2tan^-1(x))
Just like the previous problem, draw a triangle. Remember that tan^-1(x) is the same as tan(θ) = x. Draw your right triangle, the opposite side will be x, the adjacent side will be 1, and the hypotenuse will be √(x^2 + 1). Since tan^-1(x) is the same as θ, you can think of tan(2tan^-1(x)) as tan(2θ). The formula for tan(2θ):
tan(2θ) = 2tan(θ)/(1 - tan^2(θ))
Based upon the triangle we have, tan(θ) = x, now just replace tan(θ) with x to get:
tan(2θ) = 2x/(1 - x^2)
Since tan(2θ) is the same as tan(2tan^-1(x)) and tan(tan^-1(x)) = x/4, you get:
x/4 = 2x/(1 - x^2)
Now just solve for x:
x(1 - x^2) = 8x
x - x^3 = 8x
x^3 - 7x = 0
x(x^2 - 7) = 0
x = 0, ±√7
Again you need to check your answers for extraneous solutions. You'll find that x = 0 is the only solution that works. I'm assuming that 1/4x meant 1/4*x or x/4 and not 1/(4x) or 1 over 4x. Now if you meant 1 over 4x you'd get:
1/(4x) = 2x/(1 - x^2)
1 - x^2 = 8x^2
9x^2 - 1 = 0
(3x - 1)(3x + 1) = 0
x = ±1/3
In this example both 1/3 and -1/3 both work. So decide which case is yours and go with that example.