A volunteer tutor wants to know HOW to solve this problem. It is not
necessary to give the answer if you have explained it so that a
student can get it...
Solve for x: 2(sin^2)x+3sinx is greater-than-or-equal-to 2, where 0 is
less-than-or-equal-to x and x is less-than-or-equal-to 2pi.

Re write the inequality as 2sin^2(x) + 3sin(x) - 2 >= 0

Factor the left hand side to obtain (2sin(x) - 1)(sin(x) + 2) >=0

There are several ways this inequality can be satisfied.

1) Both factors are negative -- but this is not possible because
sin(x) ranges between -1 and 1, so the second factor is always
positive (i.e., the quantity (sin(x) + 2) ranges between 1 and 3).

2) Both factors positive -- we've already seen that the second factor
is always positive, so we need to find x such that (2sin(x) - 1) > 0.

Rewriting this as 2sin(x) > 1, sin(x) > 1/2,  pi/6 < x < 5*pi/6

3) Either factor equal to zero -- we've already seen that the second
factor can never be zero, so we only need to ask when the first factor
is equal to zero.  That's simple, given what we already did in case 2
above.

The final solution is: pi/6 <= x <= 5*pi/6