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Solve the following equation for &on the interval [0,2r): sec (0) Choose all correct solutions from the answer choices below:Select all that apply:#15Enig equat...

Question

Solve the following equation for &on the interval [0,2r): sec (0) Choose all correct solutions from the answer choices below:Select all that apply:#15Enig equation nas no solutions

Solve the following equation for &on the interval [0,2r): sec (0) Choose all correct solutions from the answer choices below: Select all that apply: # 1 5 Enig equation nas no solutions



Answers

Solve each equation for exact solutions in the interval $0 \leq x<2 \pi$ $$\sec ^{2} x+\sqrt{3} \sec x-\sqrt{2} \sec x-\sqrt{6}=0$$

Mm. So we have an equation of salt which is second minus sinking X minus route to. Mhm. Second X minus route two is equal to zero. And we're trying to figure out what this X. Is. So the first thing we should try to do is to get the second X alone. In order to do that, we need to move that route too. So in order to move it from the left side to the right I'm going to add a route to because what I do to one side I do to the other giving me now seek an X is equal to route to. Now. There are different ways that I could solve this. The one that I like to use is by knowing so gotta knowing sine cosine and tangent in terms of my unit circle, I have my four quadrants here and I have a handy trick called All students take calculus. So all students take calculus, I'm just gonna put help to make it easier for myself. So the reason that I have this written down is because all students take calculus is telling me where I'm going to have positive or negative versions of all my terms meaning sign tangent and co sign students is saying that sign will be positive here. So I'm gonna say all for everything is positive, only sign is positive, take is going to be tangent and calculus is going to be co sign. But right now I can't use it because I am talking about second. So I need to think about their inverse relationships what pairs with second? Well, another way you can write second is one over co sign. So I have one over CO sign is now equal to route to but I don't like it there. I need to move that coast sign. So what I can do is multiply it across with that route to and do the same kind of like switching them. Give me one over root two is equal to cosine X. Another way I can write one over it too is route 2/2. Now this is something that I know I can quickly figure out co sign of X is equal to route to over to. That should be a known pair. So what I can end up doing here is actually think about all the places that are gonna have route to over two. The only one that's going to have that is with Power four. So I can write all my versions of Power four. So that be pie or 43 pi or four, £5 or four and seven pi over four. And the reason I'm only writing these four is because we are looking at from zero to two pi. Yeah, is the only places that I could be. So. First thing I need to think about is I need the positive route to or to say anything about where is co sign positive. And I use my all students take calculus in order to figure this out so I'm gonna write them. All students take calculus so I need the ones that just have co sign all house. Co sign meaning sign tangent and co sign is positive. So I'm gonna keep that s means sign is positive. Well that would mean CO sign is negative, coastlines negative. That is not my answer. T for tangent tangent is positive which means co sign and sign of negative. Again something I don't want C. Stands for co assignment and co sign is positive which is what I need. So my answer is going to be pi over four and seven pile before. Yeah.

When we're trying to solve seeking Squared X plus two seeking experts for equal zero. So we'll let s equal seek index. Then this is s squared plus two s plus four and ah, zero. It doesn't look like it. Factors nicely. So we're gonna have to solve this by the quadratic formula. Um so S is equal to negative to plus or minus square root B squared. So four minus for time's A, which is one times C, which is four all over two times a day. So that's ah, two times one. Um, this is quadratic formula. Ah, If we simplify and will be negative to plus or minus root four minus 16 is minus 12 over to then. Since we're talking about real solutions, we cannot solve with a root negative 12 Ah, in the in the equation. So there are no solutions. So this shrink equation

So given given aggression is caused five X minus cost three X. Is equal to zero for taking calls on their threats of cost five X. Is equal to cause three X. Now comparing with we know that cause tita is equal to cause as far this implied that tita is equals two and five plus someone so far. So for similarly for this five X. Z equals 22 and five plus or minus three X. So now we're having two cases. The first one is five X. Is the five X. 0 to 25 plus uh Three X. And the second case is five X. Is equal to 25 minus three X. Now first solving this part. So five that is two weeks as equals to two and five. This implies exes equals to end five. And here are a value of uh in here. Uh Eight X. Is equal to two empire. And the access it goes too. And bye bye. Four. Now our X. Lies between 0 to 2 pi uh uh pi is excluded not included and zero is included. So now putting the values of and and is and and desires of putting the values of and So x. comes out to be zero zero than five and uh two. But by the clearance or to buy won't be there from first zero. And pi came out came out to be the answer. And from here X comes out to be by by four Then 254 that is five x 2 then 354 Then 45 before that is pie. Then 5554654. That is 352, uh 75 by four and 854. That is uh again uh to pass or to buy a curious. So the final answer here, Zero and the fire also. Uh and here's zero here. Zero would also be there. So zero and five are also included over a year on leave, sort of final answer comes out to be X is equal to n pi by four. And since our exes between 0 to 2 pi, So the answer is zero. Bye bye for Bye Bye. two, 054, Zero by by four by by 23 by by food. Bye 5554, 352. Then 75 back.

We're being asked to solve out the equation two times seeking of X plus one equals seeking of X plus three. So for me when I look at this, my goal, I mean, you know if you just treat it like any normal algebra problem, if you look at this as we have two X terms in this case it's seeking of X terms. But if we say we have two X terms in two constant terms, we want to get all the X terms, all the variable terms on one side and all the constants on the other side. So in this case I want to get my two seeking terms seeking of X terms on one side and my two constants on the other side. So we'll say I want to get my constants on the right side, meaning I want to subtract one over here so that all my constants are on the same side. And I want to get my secret of X terms on the left side, which means I want to subtract this second of X term to the other side. So in that case to seeking of X minus seeking of X or if it helps, you could think of it as minus one seeking fx right? Because there's always a one in front of any, there's always a one coefficient there if we don't see anything else. So two minus one means we are left with just one single secret of X term on the left side, three minus one gives us to All right, there's different ways you can look at that. But I mean, eventually, however, you choose to solve that down, you're gonna end up with seeking of X equals two. Now we know that second of xr second is the inverse of Cosan, meaning we could rewrite seeking of X as one over co sign of X and if one over cosine of X is equal at two. That is equivalent to saying that the co sign of X is equal to one half. If that's too much of a jump just to prove why that works. If we wanted to solve this out, um the way we do that is we would want to get co sign of X out of the denominator, so we would multiply both sides by the co sign of X. So if we multiply both sides by the co sign of X, that would cancel out on the left side and we would have won equals two times the co sign of X. And then the A co sign of X by itself. We would divide by two and that's how we end up with the co sign of X equals one half. Maybe you needed that to really understand why maybe you didn't, but it didn't take that long either way. So we have the coastline of X equals one half. That's an equation that is much simpler to solve out than what we had been dealing with. We know co sign is the X coordinates on the unit circle, so we have the unit circle over here on the left side of the screen. We simply just need to find where we have positive one half as an X coordinate. That would be right here and right here. Since the variable is X, we want to use our radiant measurements so it looks like we'd be using pi over three and five pi over three. So those would be our two answers for this problem.


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