[MUSIC] Let's see an example of how to solve trig equations that are quadratic

in form. [SOUND] Let's find all solutions to this

equation here. Now, let's begin by bringing the 1 to the

left-hand side which would give us 2sine^2 theta + sine theta -1 = 0.

Now, isn't this the same thing as 2u^2 + u - 1 = 0, where u is sine of theta so

it's quadratic in form. And how do we solve a quadratic equation? Well, we

factor if we can and the left-hand side here will factor.

It factor into 2u - 1 * u + 1, which means this will factor in the same way,

namely it factors into 2sin theta - 1 * sine of theta + 1. And now, we have a

product of factors equal to zero. Which means, either this first factor,

2sine theta - 1 is zero or the second factor,

sine of theta + 1 = 0. So, let's solve each equation separately.

So, let's start with this first equation here.

So case 1, 2 * sine of theta - 1 = 0 or solving for sine of theta, we get sine of

theta = 1/2. Which means, we need to find all angles

theta whose sine is equal to 1/2. And let's recall our units circle.

[SOUND] Remember, that the sine is the y-coordinate of the point of intersection

of the terminal side of the angle and the unit circle.

So, looking ove here in Quadrant I, we see that the y-coordinate is the 1/2

which corresponds to the angle of pi / 6. But also over here in Quadrant II, the

y-coordinate is also 1/2, which corresponds to the angle of 5pi /

6. So, both of these angles will have sine =