5.5 Trigonometric Equations

Transcription

1 5.5 Trigonometric Equations Section 5.5 Notes Page You should have received a unit circle sheet. If not this is available on the website. This allows us to see the exact values of certain angles between 0 and 0 degrees. Now we don t need to use reference angles. This section will cover how to solve trigonometric equations which is one skill you will need in calculus. The main strategy is to isolate the trig function. The we will take the inverse trig function of both sides to get the answer. x on [0 0 The coe is already isolated so now we will take the inverse coe of both sides. x. Now simplify. x What we need to do now is look at the unit circle sheet and find ANY angles between 0 and 0 that give an x value of coe and y corresponds to e.. Remember x corresponds to x 0 0 Both of these will give a value of. x. x First we isolated the e. Now we need to take the inverse e of both sides. x Simplify. x We need to find ANY angles on the unit circle that give a y value of 7 x The above is not my complete answer. This problem did not give an interval like [ 0 ] to find your answers. Because of this our answers will not only be in the first revolution of the circle. If no interval is given we need to add a k to our answer. The k value represents how many times we are going around the circle until we 7 come to our answers. So we will write: x k k. We could have also written our answers in degrees as well: x 0 0 k 0 0 k..

2 x 0 on [ 0 Section 5.5 Notes Page x First we isolate the coe. Now we need to take the inverse coe of both sides. x Simplify. x We need to find ANY angles on the unit circle that give a x value of 5 x. These are my only answer ce they gave us an interval [ 0 ]. Since this interval is given in radians we must write our answers in terms of radians. x on [0 ). To solve this one take the square root of both sides: bottom separately: of x. The angles are: 5csc x on [ 0 x. You can take the square root of the top and. To solve this look on the unit circle and find all angles that have an x value 5 7. First isolate the ecant by adding to both sides and then dividing both sides by 5. csc x We will use the identity csc x. So now the problem becomes: x x Cross multiply. x Take the inverse e of both sides. x Simplify. x Look on the unit circle and find ANY angles that give a y value of. x This is the only gle value on the unit circle.

3 cot 0 on [0 0 ) Section 5.5 Notes Page First isolate the cotangent by subtracting one from both sides and then dividing both sides by. cot Now use the formula: cot. tan tan Cross multiply. tan Since there is no tangent on our unit circle look for ANY angle such that if you divide the y by x (y/x) you will get The coe is isolated so now we will take the inverse coe of both sides: Simplify. We need to find ANY angles on the unit circle that give a x value of. Since there is no interval given we need to add a 0k to our answers. 0 0 k For each of our answers we need to solve for by dividing by k 0 80 k These are our answers k ( ) on [ 0 We will proceed the same way we did the previous example. In this one we need to take the inverse e of both sides: We need to find ANY angles on the unit circle that give a x value of. This 5 will be and. On this one because there is something inside the trig

4 Section 5.5 Notes Page function that is not just theta we will also be ug the k values. Because we need to use radians we will add k. 5 k k For each of our answers we need to solve for by dividing both sides by. k 5 k These are our equations. To get the answers in our interval we will be putting values in for k. First we will start k = 0 until we get a number that is outside our interval. k 5 k When k = 0 we have: (0) so 5 When k = we have: () so If we let k = then we get something that is more than which is outside our interval so we will stop. 5 5 When k = 0 we have: (0) so 5 When k = we have: () so If we let k = then we get something that is more than which is outside our interval so we will stop. So our answers to this problem are: 5 5. ( ) 0 on [0 0 First we need to solve the above for coe: ( ). We will proceed the same way we did in the previous example. In this one we need to take the inverse coe of both sides: ( ) We need to find ANY angles on the unit circle that give an x value of. This will be 5 and 5 degrees. Now we set up our two equations as before. 5 0 k 5 0 k For each of our answers we divide both sides by. 5 0 k 05 0 k These are our equations. To get the answers in our interval we will be putting values in for k. First we start with k = 0 until we get an angle that is outside our interval.

5 5 0 k When k = 0 we have 5 0 (0) so 5. When k = we have 5 0 () so 5. Section 5.5 Notes Page 5 When k = we have 5 0 () so 55. If we let k = then we get an angle that is more than 0 degrees which is outside our interval so we will stop k When k = 0 we have 05 0 (0) so 05. When k = we have 05 0 () so 5. When k = we have 05 0 () so 5. If we let k = then we get an angle that is more than 0 degrees which is outside our interval so we will stop. So our answers to this problem are: EXAMPLE: Solve the equation: x x 0 on [ 0 For this one the only thing we can do is factor out the common factor which is x : x ( x ) 0. Now we need to set each factor equal to zero. We will get x 0 and x 0. We need to solve each equation separately. For the equation x 0 we need to look at the unit circle where the x value is zero. This will happen at and at. For the second equation x 0 this is the same as x. Again we look at the unit circle and find all places where the x value is. We will get 0. The angle of zero radians is also the same as radians but we don t write the because that is not included on our interval. EXAMPLE: Solve the equation: x x 0 on [ 0 We can factor this one: ( x )( x ) 0. Now set each part equal to zero. We get x 0 and x 0. Solving the first equation we will get x. We look on the unit circle and look for any x values that are 5. This will happen at and. Solving the second equation you will get x. The 5 angle that gives an x value of negative one is. Therefore our answers are and. EXAMPLE: Solve the equation: x x x on [0 0 We need to set this equal to zero: x x x 0. Now factor out a common factor of x : x( x ) 0. Now set each factor equal to zero. We have x 0. Looking at our unit circle we see that x is 0 and 80. The other equation gives us x. Taking the square root we get x. This means that x.. Since this number is larger than one this will not give us any solutions because of the domain of the coe. So our answers for x are 0 and 80.

6 EXAMPLE: Solve the equation: tan x tan x tan x 0 on [ 0 Section 5.5 Notes Page This one can be factored by grouping. I will factor out a tan x from the first two terms and then a negative from the second two terms. This will give us: tan x(tan x ) (tan x ) 0. There is a common factor of tan x that I will factor out: (tan x )(tan x ) 0. Now we set both factors individually equal to zero. 5 For the first equation we have tan x 0 in which tan x so x and x from the unit circle. For the second equation we have tan x 0 in which tan x. Since we get a positive or a negative this means that we will get an answer in all four quadrants of the unit circle. We get the following: Our final answer if all of the following angles:. EXAMPLE: Solve the equation: 0 on [0 0 Since there are both es and coes we need to use an identity to get all the terms to have the same trig value. Since I notice there is already a e in the problem I want to use. So now the problem is: ( ) 0. Simplifying we get: 0. When we solve this we get so. Then values off the unit circle are: Notice that ce our interval was given in degrees we can write our answers in degrees. EXAMPLE: Solve the equation: on [ 0 We need to use another identity on this one. This time we will use. So now our problem becomes:. Setting it equal to zero will give us: 0. We can factor out a coe to get: ( ) 0. Setting the first term equal to zero will give us 0 so we know from the unit circle. Setting the second term equal to zero we will get 0 so 5 5. Then we know from the unit circle. So our final answer is:. EXAMPLE: Solve the equation: cot x csc x 0 on [0 0 Once again we need to use a trig identity so that all of these are in terms of the same trig value. We can either use all cotangents or all ecants. I will use the identity cot x csc x : (csc x ) csc x 0. Simplifying we get: csc x 0. Solving for ecant we get csc x. I will now use the identity csc x. This will change the problem into:. Cross multiplying gives us x so x x x. Solving for e we get x. Then we know from the unit circle: x