Help with precal homework please

2 sin(x) + 3 > sin²(x)

0 > sin²(x) - 2 sin(x) - 3

0 > (sin(x) - 3) (sin(x) + 1)

The right side is 0 when either

sin(x) - 3 = 0   or   sin(x) + 1 = 0

sin(x) = 3  or   sin(x) = -1

The first case offers no real solutions, so we're left with

sin(x) = -1

which happens for

x = -π/2 + 2nπ

where n is any integer.

In the interval 0 ≤ x ≤ 2π, the expression is 0 for x = 3π/2 (when n = 1). So check the sign of the expression when x is picked from two different intervals:

• If 0 ≤ x < 3π/2, then sin(x) - 3 < 0 and sin(x) + 1 > 0, so their product sin²(x) - 2 sin(x) - 3 < 0.

• If 3π/2 < x ≤ 2π, then sin(x) - 3 < 0 and sin(x) + 1 > 0, so the product is again negative.

In both cases, sin²(x) - 2 sin(x) - 3 < 0, so we have the solution set

0 ≤ x < 3π/2   and   3π/2 < x ≤ 2π

Another way to arrive at the same conclusion:

0 > sin²(x) - 2 sin(x) - 3

Recall the half-angle identity for sin:

sin²(x) = (1 - cos(2x))/2

So we have

0 > (1 - cos(2x))/2 - 2 sin(x) - 3

0 > -cos(2x) - 4 sin(x) - 7

0 < cos(2x) + 4 sin(x) + 7

Both sin(x) and cos(x) are bounded between -1 and 1. If they are both minimized, so that cos(2x) = sin(x) = -1, then the right side is at least

-1 + 4(-1) + 7 = 2

so the inequality holds for all x in the interval, except for x = 3π/2.

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.