Find all solutions to the equation where 0 ≤ θ\thetaθ ≤ 2π2\pi2π.sin⁡θcos⁡(2θ)−sin⁡(2θ)cos⁡θ=22\sin\theta\cos\left(2\theta\right)-\sin\left(2\theta\right)\cos\theta=\frac{\sqrt2}{2}sinθcos(2θ)−sin(2θ)cosθ=22

θ = 5 π 4 , 7 π 4 \theta=\frac{5\pi}{4},\frac{7\pi}{4} θ = 4 5 π , 4 7 π

Find all solutions to the equation where 0 ≤ θ \theta θ ≤ 2 π 2\pi 2 π . sin ⁡ θ cos ⁡ ( 2 θ ) − sin ⁡ ( 2 θ ) cos ⁡ θ = 2 2 \sin\theta\cos\left(2\theta\right)-\sin\left(2\theta\right)\cos\theta=\frac{\sqrt2}{2} sin θ cos ( 2 θ ) − sin ( 2 θ ) cos θ = 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

In this problem we're asked to find all solutions for the equation between 0 and 2 pi and the equation that we're given here is the sine of theta times the cosine of 2 theta minus the sine of 2 theta times the cosine of theta is equal to the square root of 2 over 2. So let's dive right into problem. Now your first inclination might be to go ahead and expand these two terms out using your double angle identities, But that's actually not going to help us here. Let's take a deeper look at this left side. I'm taking the sine of theta times the cosine of 2 theta and then subtracting the sine of 2 theta times the cosine of theta. Now since I have a sine and a cosine and then I'm subtracting another sine and a cosine, this reminds me of my sum and difference identities for sine. And because I'm subtracting here, it's specifically my difference formula for sine. So I can actually rewrite this using that identity to get the sine of theta minus 2 theta. Now, you could go ahead and expand this out to double check that you get this same exact thing, but you will. So now all we have is the sine of theta minus 2 theta is equal to the square root of 2 over 2. Now what is theta minus 2 theta? It's negative theta. So this is the sign of negative theta is equal to root 2 over 2. But we never want to have a negative argument. Right? So we can go ahead and use our even odd identities here in order to write this sine negative theta as negative sine of theta. Now I have negative sine of theta is equal to root 2 over 2 and all that's left to do is solve the resulting linear equation. Now the first thing I need to do here is isolate my trig function by getting rid of this negative. So if I divide both sides by negative one, I end up with the sine of theta is equal to negative root 2 over 2. And now I just need to think of my angles on the unit circle for which the sine is equal to negative root 2 over 2. Now those angles for which that is true is theta equals 5 pi over 4 and theta equals 7 pi over 4. Now you might be tempted here to go ahead and add 2pi into each of these solutions. But in our problem, we're specifically asked to find solutions between 0 and 2pi, so just angles on our unit circle, so we're actually done here and this is our final solution set. Thanks for watching and I'll see you in the next one.