6.5 Trigonometric Equations

Transcription

1 6. Trigonometric Equations In this section, we discuss conditional trigonometric equations, that is, equations involving trigonometric functions that are satisfied only by some values of the variable (or possibly, are not satisfied by any values of the variable. The values that satisfy the equation are called solutions of the equation. Eample 1: Solve the equation: cos θ = a. over [0,. b. over the set of real numbers. ( a. On [0,, there are only two angles that satisfy the equation. Since cos = ( 6 11 cos = 6, the solutions over [0, are θ = 11 and θ = 6 6. b. Since the cosine function has period, all solutions of cos θ = may be given by the general formula where k is any integer. θ = 6 Eample : Solve sin + = 0 a. over [0,. b. over the set of real numbers. + k and θ = k a. First isolate sin : sin + = 0 = sin = = = 4, = 4, 1

2 b. Since the sine function has period, the general solution set is = 4 + k, = + k. where k is an integer. This can also be written as (4 + 6k ( + 6k =, =. Eample : ( Solve tan = 1. Note that the period of the tangent function is. In the interval [0,, the tangent function has value 1 when the argument is. Since the argument is in the 4 given equation, write the general formula that gives all the solutions. = 4 + k = = ( 4 + k ( + 4k or = 4 Therefore, the general solution set is = (or 4 + k ( + 4k = 4 Note that we could use this formula to find two solutions on [0,. To do this, we will need to plug in appropriate values for k and then simplify. For k = 0 and k = 1, we have k = 0 : = 4 + (0 = 4 k = 1 : = 4 + (1 = 7 4 Eample 4: Solve sin( = 1 on [0,. In the interval [0,, the sine function equals 1 at and, which means that 6 6 must equal and. However, here s the problem. The period of y = sin( is =. 6 6 Thus, on [0,, the graph of y = sin( will complete two cycles and will intersect the graph of y = 1 four times. Hence, the equation sin( = 1 has four solutions on [0,. General = 6 + k, = 1 + k, = 6 + k = 1 + k

3 which can be written as = (1 + 1k, = 1 ( + 1k 1 Let s plug in k = 1, 0, 1, to obtain a few solutions: k = 1 : = k = 0 : = k = 1 : = k = : = [1 + 1( 1] 1 [1 + 1(0] 1 [1 + 1(1] 1 [1 + 1(] 1 = 11 [ + 1( 1], k = 1 : = 1 1 = [ + 1(0], k = 0 : = 1 1 = 1 [ + 1(1], k = 1 : = 1 1 = [ + 1(], k = : = 1 1 = 7 1 = 1 = 17 1 = 9 1 The solutions of the given equation that lie on [0, are those that correspond to k = 0 and k = 1 (in both formulas. Hence, the solution set is 1, 1, 1 1, Eample : ( Solve 1 + cos = 0. a. Write the solution set for the general solution. b. Write the solution set on the interval [0,. a. First isolate cos ( : ( ( 1 + cos = 0 = cos = 1 Therefore, = + k, = 4 + k Multiplying both sides of each formula by yields = 4 + 4k, = 8 + 4k

4 Thus, the general solution set is = 4 + 4k, = 8 + 4k which can be written as (4 + 1k =, = (8 + 1k. b. According to the general formulas, only the solution corresponding to k = 0 in the first formula lies on [0,, i.e. = 4. 4 Eample 6: Solve sin sin + 1 = 0 on [0,. This equation is quadratic in sin and can be factored. Let u = sin. Then sin sin + 1 = 0 = u u + 1 = 0 = (u 1(u 1 = 0 = u 1 = 0 or u 1 = 0 = u = 1 or u = 1 = sin = 1 or sin = 1 Solving each equation on [0,, we obtain = 6, = 6, = 6,, 6 When a trigonometric equation contains more than one trigonometric function, identities can sometimes be used to obtain an equivalent equation that involves only one trigonometric function. (See the net eample. Eample 7: Solve cos θ + = sin θ on [0,. 4

5 This equation contains a sine and a cosine (two different functions. However, with the use of the Pythagorean Identity: sin θ + cos θ = 1, we can obtain an equivalent equation that contains only cosines. cos θ + = sin θ = cos θ + = (1 cos θ = cos θ + = cos θ = cos θ + cos θ + 1 = 0 = ( cos θ + 1(cos θ + 1 = 0 = cos θ + 1 = 0 or cos θ + 1 = 0 = cos θ = 1 or cos θ = 1 Solving each equation on [0,, we obtain θ =, θ = 4, θ =,, 4 Eample 8: Solve csc 1 = cot on [0,. csc 1 = cot = cot = cot (csc 1 = cot = cot cot = 0 = cot (cot 1 = 0 = cot = 0 or cot 1 = 0 = cot = 0 or cot = 1 Solving each equation on [0,, we obtain =, =, = 4, = 4 4,, 4,

6 When faced with a trigonometric equation that seems difficult to solve in its original form, it is natural to write the equation in a different form to make the solving process easier. To do this, we rely on algebraic techniques and the fundamental trigonometric identities that we ve learned. Eample 9: Solve sin( + sin(4 = 0 on [0,. Use the sum-to-product identity. ( ( sin( + sin(4 = 0 = sin cos = 0 = sin( cos( = 0 = sin( cos = 0 ( cos( = cos = sin( = 0 or cos = 0 = = k or = (k + 1 = = k or = (k + 1 For the first formula, we have the following solutions (corresponding to k = 0, 1,,, 4, : = 0, =, =, =, = 4, = For the second formula, we have the following solutions (corresponding to k = 0, 1: =, = 0,,,,,, 4, Sometimes squaring both sides of an equation is a good approach to try especially if it leads to an equation that is much easier to solve. However, whenever we try this approach, we must always remember to check all potential solutions since it is possible that some (or all of them could be etraneous. Eample 10: Solve tan( + 1 = sec( on [0,. 6

7 tan( + 1 = sec( = [tan( + 1] = [sec(] = tan ( + tan( + 1 = sec ( = tan ( + tan( + 1 = tan ( + 1 = tan( = 0 = tan( = 0 = = k = = k The values of that lie on [0, are those that correspond to k = 0, 1,,, i.e. = 0, =, =, =. Let s check plug each potential solution back into the original equation: = 0 : tan( tan(0 + 1 = 1 = ( : tan + 1 tan( + 1 = 1 = sec( 0 sec(0 = 1 ( = sec sec( = 1 True False = : tan( + 1 tan( + 1 = 1 = ( : tan + 1 tan( + 1 = 1 = sec( sec( = 1 ( = sec sec( = 1 True False We see that = and = are etraneous solutions. Thus, 0, We now look at the case in which a trigonometric equation has solutions that cannot be epressed in terms of the special angles. In this situation, we would need to rely on inverse trigonometric functions to write the solution set. 7

8 Eample 11: Solve the equation sin =. As usual, to solve this equation, we ask ourselves the question: What angles have a sine value of? Clearly, the solution set does not contain any of the special angles and those that are coterminal to them. However, since the range of y = sin is [ 1, 1], we know that this equation has solutions (infinitely many of them. In this case, we would need to epress the solution set in terms of the inverse sine function. From our study of inverse trigonometric functions, we know that sin = ( [ = = sin 1 where lies in 0, ]. Since this solution lies in Quadrant I, all solutions that lie in Quadrant I are given by ( = sin 1 + k. Note that sin is positive in both Quadrants I and II. Hence, we also need to find the solutions that lie in Quadrant II. To find these solutions, we use the fact that the reference angle of sin 1( is itself. This means that the solution that lies in < < is = sin 1(. Thus, the solutions that lie in Quadrant II are given by [ ( ] = sin 1 + k It follows that the solution set is ( = sin 1 + k, = [ ( ] sin 1 + k Eample 1: Solve cos = in [0,. Recall that the cosine function is negative in Quadrants II and III. To find one solution of the above equation, we can simply epress the equation in terms of the inverse cosine function: cos = ( = = cos 1 [ ] where lies in,. [ We now need to find the solution that lies in, ]. We use the fact that the reference angle of cos 1( ( is cos 1. Therefore, the solution in Quadrant III that we seek is ( = + cos 1. 8

9 It follows that the solution set on [0, is ( = ( cos 1, = + cos 1 Eample 1: Solve tan =. on [0,. Recall that the tangent function is negative in Quadrants II and IV. To find one solution of the above equation, we can simply epress the equation in terms of the inverse tangent function: tan =. = = tan 1 (. where lies on (, 0. Note that this solution is not contained in [0,. Therefore, to obtain the solution that lies < <, we will need to add, i.e. on = tan 1 (. + Since the other solution that we seek lies in < < (Quadrant II and the period of the tangent function is, we can simply add to tan 1 (. to obtain it: = tan 1 (. + It follows that the solution set is = tan 1 (. +, = tan 1 (. + An alternative approach to finding the solution set is to first use the fact that the reference angle of tan 1 (. is tan 1 (. (which lies on (0,. Therefore, the solution that lies in Quadrant II is = tan 1 (. Note that this is equivalent to the solution in Quadrant II that we obtained using the first approach. Net, since the period of the tangent function is, we add to our Quadrant II solution to obtain the solution in Quadrant IV: which simplifies to = [ tan 1 (.] + = tan 1 (. 9