grief said:Let me rearrange that a little for you:
1/2sin(2x)>sin^2(x)
which is actually nicer in the following form:
sin(x)cos(x)>sin^2(x).
This should be easier to solve.
The first step in solving this inequality is to distribute the cos^2(x) term, which gives us 1 + tg(x)*cos^2(x) > 1.
Yes, since cos^2(x) is always positive, we can divide both sides of the inequality by it without changing the direction of the inequality.
To solve for tg(x), we need to isolate it on one side of the inequality. In this case, we can subtract 1 from both sides to get tg(x)*cos^2(x) > 0. Then, we can divide both sides by cos^2(x) to get tg(x) > 0.
The solution set for this inequality is all values of x where tg(x) is greater than 0, or in other words, all values of x where the tangent function is positive.
To graph this inequality, we can plot the points where tg(x) is equal to 0, and then shade the region where tg(x) is positive. This will result in a graph with a vertical asymptote at x = 0, and a shaded region to the right of the asymptote.