1) Sin^-1 (sin (7π / 3)
Answer should be π/3
Please show the steps between.
2) Solve each inequality for x
1 < e^(3x-1) < 2
3) 1 - 2 lnx < 3
*I got x = -e but I don't know if that's right
Answer should be π/3
Please show the steps between.
2) Solve each inequality for x
1 < e^(3x-1) < 2
3) 1 - 2 lnx < 3
*I got x = -e but I don't know if that's right
-
1) The function sin^-1 always returns an angle between -π/2 and π/2. Also, for x between -π/2 and π/2:
sin^-1(sin(x)) = x
This angle, of course, is not in this range.so the answer is not 7π/3. But, notice that:
sin(7π/3) = sin(π/3 + 2π) = sin(π/3)
sin^-1(sin(7π/3)) = sin^-1(sin(π/3)) = π/3
2) Split this into two inequalities:
1 < e^(3x - 1) AND e^(3x - 1) < 2
Take the natural log of both sides. Since the natural logarithm is an increasing function, no change of sign is necessary:
ln(1) < 3x - 1 AND 3x - 1 < ln(2)
0 < 3x - 1 AND 3x - 1 < ln(2)
1 < 3x AND 3x < ln(2) + 1
1/3 < x AND x < (ln(2) + 1) / 3
Recombining:
1/3 < x < (ln(2) + 1) / 3
3) In this case, I can tell you this is not right. For one, x = -e is not even in the domain, since ln(-e) is undefined (logarithms are defined for strictly positive numbers). I suspect you might have meant x = 1/e, which is the unique x such that:
1 - 2ln(x) = 3
but this is not an equation, rather an inequality. We should expect a range of values. Solving:
1 - 2ln(x) < 3
1 < 3 + 2ln(x)
-2 < 2ln(x)
-1 < ln(x)
Taking e to the power of each side is an increasing function, so no sign change is necessary:
e^-1 < x
x > 1/e
Hope that helps!
sin^-1(sin(x)) = x
This angle, of course, is not in this range.so the answer is not 7π/3. But, notice that:
sin(7π/3) = sin(π/3 + 2π) = sin(π/3)
sin^-1(sin(7π/3)) = sin^-1(sin(π/3)) = π/3
2) Split this into two inequalities:
1 < e^(3x - 1) AND e^(3x - 1) < 2
Take the natural log of both sides. Since the natural logarithm is an increasing function, no change of sign is necessary:
ln(1) < 3x - 1 AND 3x - 1 < ln(2)
0 < 3x - 1 AND 3x - 1 < ln(2)
1 < 3x AND 3x < ln(2) + 1
1/3 < x AND x < (ln(2) + 1) / 3
Recombining:
1/3 < x < (ln(2) + 1) / 3
3) In this case, I can tell you this is not right. For one, x = -e is not even in the domain, since ln(-e) is undefined (logarithms are defined for strictly positive numbers). I suspect you might have meant x = 1/e, which is the unique x such that:
1 - 2ln(x) = 3
but this is not an equation, rather an inequality. We should expect a range of values. Solving:
1 - 2ln(x) < 3
1 < 3 + 2ln(x)
-2 < 2ln(x)
-1 < ln(x)
Taking e to the power of each side is an increasing function, so no sign change is necessary:
e^-1 < x
x > 1/e
Hope that helps!
-
Sin^-1 (sin (7π / 3) = the angle whose sine is equal to the sine of 7π/3
The angle whose sine is equal to the sine of 7π/3 is obviously 7π/3 with is coterminal in the interval of 0 to 2π with 7π/3 - 2π = 7π/3 - 6π/3 = π/3
The angle whose sine is equal to the sine of 7π/3 is obviously 7π/3 with is coterminal in the interval of 0 to 2π with 7π/3 - 2π = 7π/3 - 6π/3 = π/3