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Interactive video lesson plan for: solving trigonometric equations unit circle how to trigonometry algebra 2

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Good day students in this clip were going to be going over for examples on how to solve trigonometric equations using the unit circle. So another to get started down for this examples, you need to have on your unit circle. I have a virtual copy of my of unit circle here that of the consulting on during the problem-solving process. Right so let's go ahead and take a look at the first question so the instructions are for us to solve the following is the unit circle for of data between of zero and to pie okay said is the restriction that they are in radiant measure that means that our solution will be in radiant measures. Also okay so let's go ahead and write down question now number one for number one we are to solve negative five plus the square root of two divided by five equals negative one plus one fifth of secant data okay so the variable is solving for in this equation is paid at right so of how the software Theta to do that Tom we notice that we have fractions in our equation equation so of the five card, and denominator that the two fraction so let's go ahead and eliminates of the five to accomplish this will simply multiply both sides of our equation by five so multiply the side by five and multiply this entire equation by five to get rid of that denominator right so let's see what happened the left side is five counsel that with that side so we left with negative five plus roots to equals if you distribute the five he have negative five plus if you distribute this five to this of one fifth you have five over five which reduces to one any councils out in your left with secant theta. Right now let's go ahead and of isolate secant theta to accomplish that will add five to both sides in that leaves us with is a reflexive property of equality secant say that equals of negative five plus five councils out so you left which roots to now to get of theta isolated we're going to get rid of the secant, but what we can do here is express this trick function using one of the parents trick functions now recall that of secant theta is equal to one over cosigned Theta right so if I apply that to the left side if I express secant of as cosine theta you have to reciprocate this expression right here so this is one here this to become one over roots to okay so you reciprocate secant you end up with of cosine now of we have one over roots to hear to get rid of the radical and the denominator you rationalize were to read to that gives this of roots to over to okay so the have cosine Theta equals roots to over to now to isolate theta were going to take the inverse cosine of both sides of the equation so we have five cosine theta is equal to roots to over to we going to do the inverse cosine of the left side inverse cosine of cosine Theta in the will take the inverse cosine of the right side also inverse cosine of roots to over to face cancels out so we have to pay that equals the inverse cosine of roots to over to now how do we solve this using our unit circle your unit circle ratios is go ahead and consult how unit circle ratios of we have cosine is the x-coordinate okay cosine Theta is the x-coordinate so in this equation right here of this is asking us what x-coordinate on the unit circle has a value of roots to over to so what x-coordinate is equal to roots to over to let's write that on the side what x-coordinate is roots to over to since we recall that cosine Theta is the x-coordinate okay so let's go back to her unit circle of an end take a look at all the X coordinates that we had here which one has roots to over to so notice we have the first one here pile

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