Help with solving secxtanx=4sinx

I do not understand this problem; if you know the answer, a step-by-step explanation would be greatly appreciated!

Thanks in advance.

remember:
sec(A) = 1/cos(A)
tan(A) = sin(A)/cos(A)

So:
sec(x)tan(x) = 4sin(x)
1/cos(x) sin(x)/cos(x) = 4sin(x) [Using our identities above]
sin(x)/cos²(x) = 4sin(x) [Simplifying the fractions]
1/cos²(x) = 4 [The two sin(x)'s cancel out]
cos²(x) 1/4
cos(x) = 1/2 or cos(x) = -1/2
x = 60˚ or x = -60˚

We're trying to work out x.
So you should know that
sec(x) = 1/(cos(x)
tan(x) = sin(x)/cos(x)
Question:sec(x)tan(x) = 4sin(x)

1)
We replace the sec(x) and the tan(x), giving us ( 1/cos(x) ) * ( sin(x)/cos(x) )

2)
We multiply the numerators 1 and sin(x), then the denominators cos(x) and cos(x) which equals sin(x) / [cos(x)]^2 leaving us with
( sin(x) / (cos(x))^2 ) = 4sin(x)

3)
If we simplify this by dividing both sides by sin(x), we get:
1 / (cos(x))^2 = 4.

4) We multiply both sides by (cos(x))^2, giving us:
1 = 4[cos(x)]^2

5) Now we divide both sides by 4, which leaves us with:
1/4 = [cos(x)]^2

6) We find the square roots of both sides:
±(1/(√2)) = ±cos(x)

7) And that leaves the four following solutions:
1. cos(x) = 1/(√2) so arccos(1/(√2)) = x = π/4
2. cos(x) = -1/(√2) so arcos(-1/(√2)) = x = (3π)/4
3. -cos(x) = 1/(√2) so -arccos(1/(√2)) = x = -π/4
4. -cos(x) = -1/(√2) so -arccos(1/(√2)) = x = (-3π)/4

sec = 1/cos, tan = sin/cos
so the equation is equivalent to 1/cos^2 = 4
or cos = +/- sqrt(2)/2
the solution is x =npi +/- pi/4