Inverse trigonomtric functions arcsin 3t > pi/6
You need to substitute `arcsin (0.5)` for `pi/6` , to the right side, such that:
`arcsin (3t) > arcsin (0.5) `
You need to remember that the monotony of the arcsine function tells that it increases over R, using the definition of an increasing function yields:
`x_1 < x_2 => f(x_1) < f(x_2)`
Reasoning by analogy, yields:
`arcsin (0.5) < arcsin (3t) => 0.5 < 3t => (0.5)/3 < t`
Hence, evaluating the given inequality yields that `t in ((0.5)/3,oo).`
From the information you've given, I suppose that you want to determine the interval of values of t, for the inequality to hold.
We consider the inverse function arcsin 3t > pi/6.
We'll apply sine functin both sides:
sin (arcsin 3t) > sin pi/6
Since the sine function is increasing, we'll keep the direction of the inequality:
3t > 1/2
We'll divide by 3 both sides:
The interval of admissible values of t, for the inequality to hold, is: (1/6 ; +infinite).