Solving Trigonometric Equations Using Identities - Video Tutorials & Practice Problems

Solve Trig Equations Using Identity Substitutions

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 wanna 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 C can't squared of it minus one over the tangent of it is equal to one. I don't know how to do that. Just 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 gonna walk you through exactly how to do that. Let's go ahead and get started. Now, here we have the equation. The C can't squared of theta minus one over the tangent of data is equal to one. Now seeing that I have multiple trig functions here tells me I need to go ahead and use my trig identities in order to rewrite this in terms of just one trick function. Now, we're going to keep using our strategies for simplifying here and scanning this for identities I see in my numerator, this C can't squared of theta minus one. Now, whenever I have a squared trick function, I should be thinking about using my Pythagorean identities here. Now seeing that I have this C can't squared of theta minus one, I can use my Pythagorean identity to rewrite this giving that this is just the tangent squared of data over the tangent of data is equal to one having used that Pythagorean identity. Now, from here, I can cancel some stuff because that tangent in the bottom will cancel with one of my tangents in the top leaving me with just the tangent of data is equal to one. So we started with multiple trig functions and then we used identities in order to get down to just one trick function. And from here, we can just find our angles the 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 sine of two, the over the cosine of negative the is equal to one. Now, here, I wanna be constantly scanning for identities and something that you might notice here is that we have a sign over a cosine. So your first instinct might be to say OK, wouldn't that just be the tangent? But we can't actually do that here because these arguments are not the same In my numerator, I have two the in my argument and in my denominator, I have the negative the, so I can't use that identity. I need to think of another way to simplify this. Well, looking at that numerator, I have the sign of two theta. So having two the in that argument, what identity should we be using? Well, here we can rewrite the top using a double angle identity since we have that two theta in the argument. So using that identity here, I end up with two times the sine of data, times the cosine of data over the cosine of negative data. Keeping that the same for now is equal to one having rewritten the top there. Now looking at this, I see in my denominator, I have this negative data. So how do I get rid of that negative data? Because I don't want a negative argument here. Well, whenever we have a negative argument we want, think about using our even odd identities. So here I can rewrite the bottom of this fraction using an even odd identity. Now, I know that cosine is an even function. So I can rewrite that denominator as the cosine of data. Now everything else is staying the same for now two times the sine of data times the cosine of data and all of this is equal to one. Now, what's going to happen here? Well, I see I have cosine in the top, cosine in the bottom. So those just cancel out. Now all I'm left with is two times the sign of theta is equal to one. So I have just one trig function now and now all that's left to do is solve this resulting linear trig equation. So let's go ahead and do that. Now, I want to go ahead and isolate this trig function. So I'm going to divide both sides by two. Now that will cancel on that left side leaving me with the sign of data is equal to one half. Now, from here, I just want to think of my angles theta for which the sign is equal to one half on my unit circle. Now, the angles for which that are true is theta is equal to pi over six and theta equals five pi over six. Those are my two angles on the unit circle for which the sign is equal to one half. But remember that this problem asked us to find all solutions. So we need to go ahead and add two pi N to each of these. Now this represents all solutions to my original equation and this is my final answer here. Theta is equal to pi over six plus two pi N and theta is equal to five pi over six plus two pi N. Now that we know how to use identities to solve some more complicated trigger equations. Let's continue practicing together. Thanks for watching and I'll see you in the next one.

Find all solutions to the equation.

$\left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12$

• A. $\theta=\frac{5\pi}{12}+2\pi n,\frac{7\pi}{12}+2\pi n$
• B. $\theta=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n$
• C. $\theta=\frac{\pi}{3}+2\pi n,\frac{2\pi}{3}+2\pi n$
• D. $\theta=\frac{\pi}{3}+\pi n,\frac{2\pi}{3}+\pi n$

Find all solutions to the equation where 0 ≤ $\theta$ ≤ $2\pi$.

$\sin\theta\cos\left(2\theta\right)-\sin\left(2\theta\right)\cos\theta=\frac{\sqrt2}{2}$

• A. $\theta=\frac{5\pi}{4},\frac{7\pi}{4}$
• B. $\theta=\frac{\pi}{4},\frac{3\pi}{4}$
• C. $\theta=\frac{5\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n$
• D. $\theta=\frac{3\pi}{4},\frac{7\pi}{4}$