Hey, everyone. When solving linear trig equations, we wanted to find an angle theta that made our equation true. And now that we're tasked with solving some more complicated trig equations, we still want to do the same thing. We want to find an angle theta that makes our equation true. But now that these equations have more than just one trig function, it's not quite so simple because how are we going to find an angle theta for which the secant squared of it minus 1 over the tangent of it is equal to 1? I don't know how to do that just using the unit circle. But now that we know a bunch of different trig identities, we can now use those identities in order to rewrite these equations in terms of just one trig function. Then we're just left with a linear trig equation that we already know how to solve. So here, I'm going to walk you through exactly how to do that. Let's go ahead and get started.

Now, here we have the equation the secθ2-1secθ²2-1tanθ=1. Seeing that I have multiple trig functions here tells me that I need to go ahead and use my trig identities in order to rewrite this in terms of just one trig function. Using the Pythagorean identity, we can rewrite this, giving that this is just tanθ2/tanθ=1. Now from here, I can cancel because one tangent in the numerator will cancel with the tangent in the denominator, leaving me with just the tangent of theta is equal to 1.

So, we started with multiple trig functions and then used identities to get down to just one trig function. From here, we can find our angles theta on the unit circle for which this is true and then add this factor of pi in in order to account for all possible solutions. Now not all of these equations are going to be quite so simple and we're not always just going to use our Pythagorean identities. So let's go ahead and walk through another example together.

Here, we're asked to find all solutions to the equation sin(2θ)cos(-θ)=1. Now, note that we have a sine over a cosine, but these arguments are not the same. In the numerator, I have 2θ, and in the denominator, I have -θ. We can't directly use the tangent identity here, so we need to think of another way to simplify this. Using the double angle identity, we can rewrite the top as 2·sinθ·cosθ over the cosine of negative theta. Since cosine is an even function, this simplifies further to 2 times the sine of theta equals 1. Now, I just need to solve this resulting linear trig equation.

By dividing both sides by 2, we get the sine of theta equals 1/2. The angles for which this is true are theta equals pi/6 and theta equals 5pi/6. But we need to find all solutions, so we'll add 2pi n to each of these. Thus, theta equals pi/6 + 2pi n and theta equals 5pi/6 + 2pi n. Now that we know how to use identities to solve some more complicated trig equations, let's continue practicing together. Thanks for watching, and I'll see you in the next one.