- All right, here, we're going to make use of this idea of reducing a power to solve a trig equation. So suppose we have cosecant to the fourth of 2x equals 4. Well, again, let's try to reduce the exponent here. So if we take the square root of both sides, cosecant to the fourth of 2x-- if we take the square root of the left side, well, I'm also going to take the square root of the right side, put my positive and negative over there. 

Well, if you take something to the fourth power and you square root it, it's going to give us something to the second power. So that's simply going to give us cosecant squared of 2x equals positive or negative 2. And I'm actually, I think, going to do this one more time because now we can get rid of the fact that it's being squared by taking yet another square root, so cosecant squared of 2x. Notice in this case, since we're going to take the square root, the fact that it equals negative 2, there would be no solutions to this equation. 

So really, we should be a little more careful. We should think at this point there's really two equations, cosecant squared of 2x equals positive 2, which could have a solution, and we have cosecant squared of 2x equals negative 2, but you can't square something and get a negative out at the end. So really, the only equation that makes sense here, the only equation that would have solutions would be if it equals positive 2. So definitely keep that in mind. 

So OK, we've got positive 2. Again, I'm square rooting both sides. And again, I have to include this positive and this negative. All right, so now I'm just going to be left with cosecant of 2x equals positive or negative square root of 2. And now what I'm going to do is, well, cosecant, that's really the same thing as 1 over sine. So we'll have 1 over sine of 2x equals positive/negative square root of 2. 

What I'm going to do next is you could multiply both sides by the sine of 2x. So if we multiply both sides by sine of 2x and by sine of 2x, we would be left with a 1 on the left side. We'd have positive/negative square root of 2 times sine of 2x. And now I'm just going to divide both sides by my positive/negative root 2. 

So we would have 1 over positive or negative square root of 2 equals sine of 2x. And we can always rationalize this denominator. If we multiply top and bottom by square root of 2, we'll get square root of 2 on top. Square root of 2 times the square root of 2 will simply be 2, plus or minus. So really, all of this reduces to solving the equation. Sine of 2x equals positive or negative square root of 2 over 2. 

Again, we wanted our solutions to be in the interval 0 to pi. So OK, now we simply get two equations. We've got sine of 2x equals the positive square root of 2 over 2. And then we have sine of 2x equals negative square root of 2 over 2. 

And I'm going to do this idea that I think we've seen in some other videos I'm going to replace the 2x. I'm going to let 2x be replaced with y. And again, since x was in the interval 0 to pi, 2x would be in the interval 0 to 2 pi. So really, I'm going to find solutions. I'm going to let sine of y equal positive square root of 2 over 2. 

And since I want 2x to be in the interval 0 to 2 pi, equivalently, that means I can find solutions to this equation where y is in the interval 0 to 2 pi. Well, in the interval 0 to 2 pi, sine equals square root of 2 over 2 at the angle pi over 4. I'll do everything in radians, pi over 4, and also at 3 pi over 4. 

We could do the same thing with our other equation, just replace our 2x with y. Well, sine of y equals negative root 2 over 2. Well, this is going to happen in quadrant 3 and 4. And that's going to be the angle 5 pi over 4. Whoops, I wrote x. This should definitely be a y here, so used to writing x's. So y would equal 5 pi over 4. Or it could also equal 7 pi over 4. 

So now we know what y would equal. Well, we don't want to know about y. We want to know about x. So now we'll just simply let 2x equal pi over 4. That would give an equation we have to solve. 2x would have to equal 3 pi over 4. Likewise, 2x would have to equal 5 pi over 4. Or 2x would have to equal 7 pi over 4. 

In all cases, what we're going to do is simply multiply both sides by 1/2. So if we multiply both sides by 1/2, for our first equation, we'll just get x equals pi over 8. For our next one, we'll get x equals 3 pi over 8. Then we'll get x equals 5 pi over 8. And our last solution would be x equals 7 pi over 8.