Solving Trig Inequalities...Need help.

[Gsingh]

Hello everyone,
I looked everywhere online and nobody properly explains how to solve trig inequalities, and i'm still stuck. This is the problem...

2sin²x > sinx , 0 < x < 2pi

I got to this point...

sinx(2sinx - 1) > 0

I then split the two expressions and solved, which gave me solutions of 0, pi, pi/6, and 5pi/6. It wants sin > 0 and sin > 1/2...so the final solution I ended up with was [0, pi] U [pi/6, 5pi/6] and it seemed right to me because this is where the expressions are positive but when I checked the given answer in the back of my packet, it did not match

The actual answer is [0] U [pi/6, 5pi/6] U [pi, 2pi)

How would I get this answer? Thanks in advance for the help

[anonmeans]

2sin²x > sinx , 0 < x < 2pi

I got to this point...

sinx(2sinx - 1) > 0

I then split the two expressions and solved, which gave me solutions of 0, pi, pi/6, and 5pi/6.

I think you solved the EQUALITY, not the INequality. You did sin(x)[2sin(x) - 1] = 0 and got where the graph crosses the x-axis. Now you have to find where BOTH factors are positive or BOTH factors are negative (so the product is positive).

For that, you might want to use something like the method they use here. The sin(x) > 0 for 0 < @ < pi. ("@" is "theta".) The 2sin(x) - 1 > 0 for sin(x) > 1/2, so for pi/6 < @ < 5pi/6. Now look at your intervals:

0 < @ < pi/6: sin(x) > 0 but 2sin(x) - 1 < 0 [NO]
pi/6 < @ < 5pi/6: sin(x) > 0 and 2sin(x) - 1 > 0 [YES]

Now do the other interval. Then remember that the question is for "or equal to", so the zeroes at the ends of the x-axis intervals count, too. That's why [0] (the "interval" containing just zero) is part of the solution. You can do a graph of y = sin(x)[2sin(x) - 1] to check.

[Gsingh]

Thanks for the help, I think I can do it now. By the other interval you mean when both expressions are negative, right?