trignometry problem?

Question

If sin theta = x, ( -pi/2 < theta
16x^4 - 8 sqrt(3) x^3 - 24 x ^2 + 18x + 12sqrt(3) x -9sqrt(3) <0 ,
then which of the following best represents x ?


1) 0
2)-1
3)sqrt(3)/2
4)-1

whats the easiest way to solve this problem ?

i have to choose an answer from above .

i need a smart answer .

i think , one should not solve that big equation to get the answer .

do you find any catch in this problem ?

thank you

Answer 1

In the interval given, -1 =< sin theta =< 1.
Therefore -1 =< x =< 1
Thus 2) -1

Answer 2

16x^4 - 8 sqrt(3) x^3 - 24 x ^2 + 18x + 12sqrt(3) x -9sqrt(3)
= (16x^4 - 24 x ^2 + 18x) + (- 8 sqrt(3) x^3 + 12sqrt(3) x -9sqrt(3))
= 2x(8x^3 - 12x + 9) - sqrt(3)(8x^3 - 12x + 9)
= (8x^3 - 12x + 9)(2x - sqrt(3))

By the rational root theorem, we note that x = -3/2 is a root
of 8x^3 - 12x + 9. Then by long division,
8x^3 - 12x + 9 = (x + 3/2)(8x^2 - 12x + 6).

The discriminant of the quadratic 8x^2 - 12x + 6 is
(-12)^2 - 4*8*6 = 144 - 192 < 0, so it has no real roots.

Then
16x^4 - 8 sqrt(3) x^3 - 24 x ^2 + 18x + 12sqrt(3) x -9sqrt(3) <0
iff
(8x^3 - 12x + 9)(2x - sqrt(3)) < 0
iff
CASE 1: 8x^3 - 12x + 9 > 0 and 2x - sqrt(3) < 0
or
CASE 2: 8x^3 - 12x + 9 < 0 and 2x - sqrt(3) > 0


CASE 1)
8x^3 - 12x + 9 > 0 and 2x - sqrt(3) < 0
iff
(x + 3/2)(8x^2 - 12x + 6) > 0 and 2x - sqrt(3) < 0
iff
CASE 1.1: x + 3/2 > 0, 8x^2 - 12x + 6 > 0, and 2x - sqrt(3) < 0
or
CASE 1.2: x + 3/2 < 0, 8x^2 - 12x + 6 < 0, and 2x - sqrt(3) < 0

From CASE 1.1 we obtain -3/2 < x < sqrt(3)/2 since
8x^2 - 12x + 6 > 0 no matter what x is. Why? Its discriminant
is negative and 8 > 0, so the parabola (opens up) stays
above the x-axis always.

Note that CASE 1.2 is invalid since |x| < 1 (from the hypothesis).


CASE 2)
8x^3 - 12x + 9 < 0 and 2x - sqrt(3) > 0
iff
(x + 3/2)(8x^2 - 12x + 6) < 0 and 2x - sqrt(3) > 0
iff
CASE 2.1: x + 3/2 > 0, 8x^2 - 12x + 6 < 0, and 2x - sqrt(3) > 0
or
CASE 2.2: x + 3/2 < 0, 8x^2 - 12x + 6 > 0, and 2x - sqrt(3) > 0

From CASE 2.1 is invalid since 8x^2 - 12x + 6 > 0 no
matter what x is as explained in CASE 1.1.

Note that CASE 2.2 is invalid since |x| < 1 (from the hypothesis).


Hence, CASE 2 never holds and only result we obtain
from CASE 1 is -3/2 < x < sqrt(3)/2.

Now the hypothesis limits |x| < 1, so we need to cut the
lower bound from -3/2 to -1. Hence, the final answer is
-1 < x < sqrt(3)/2.

I don't see any smart way.


Hence, none of the answer choices are is the answer
unless there was a typo.

Answer 3

answer #2) -1
Factor the quartic function to get:

(2x - √3)(2x + 3)(4x^2 - 6x + 3) < 0

(4x^2 - 6x + 3) is always positive, since the discriminant b^2 - 4ac<0 implies no real roots and thus the curve never crosses the x-axis.

Therefore, one and only one of the remaining factors is negative for the whole quartic to be negative.
There are two cases:

1)
2x - √3 > 0 and 2x + 3 < 0
x > √3/2 and x < -3/2
which is impossible

2) 2x - √3 < 0 and 2x + 3 > 0
x < √3/2 and x > -3/2

But |x| < 1 since x = sin(theta), raising the lower bound to -1 and leading to the final answer

Answer 4

well if u dont know the trig function they are: sin=opposite over/adjacent cos=adjacent/hypotenuse tan= opposite/adjacent remember this by the acronym soh cah toa Then draw a pic of what u have and plug the things u have into the appropriate formula. If i have the right pics then u want to use sinx=7.5(opp)/9.3(hyp) Use a graphing calculator and hit the 2nd button then sin to get sine inverse and just put the 7.5/9.3 in the paranthesis

Answer 5

x = sin ( theta ) and -pi/2
this means that 0
1) represents x best