How do you solve 2sinsquared x 5sinx 3 0?

Answer 1

I presume that you are asking:

2 * sin^2(x) + 5*sin(x) + 3 = 0

This one is actually easy and you can avoid doing any tedious calculations by noticing that the range of sin(x) is [-1,1] while the range of sin^2(x) is [0,1]. Also note that every time sin(x) = -1, sin(x)^2 = (-1)^2 = 1.

Like good little Calculus students, we remember the unit circle which we memorized in high school trigonometry/advanced algebra/precalculus. The unit circle reminds us that sin(x) = -1 when x is (3/2) * pi. We also remember that sin(x) repeats itself for every 2*pi.

So our solution set is:

(3/2)*pi + 2*pi*n, where n is any integer.