Section 8.3 Trigonometric Equations


 Πύῤῥος Αναγνωστάκης
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1 99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions. Many times, an equation with a trigonometric function will have an infinite number of solutions. Thus, we will need to write the general fm of the solution. Find the a) general solution as well as b) eight particular solutions: Ex. 1 cos(x) = If we examine the graph of y = cos(x) and y = are an infinite number of points the two curves intersect:, we see that there In the interval [0, π), the cosine function is positive in the first and fourth quadrants. In the first quadrant, x = cos 1 ( ) = π. Since π is also our reference angle, then π π = π is the angle in the fourth quadrant. The cosine function is periodic with period π, so these solutions will repeat every π. Thus, our general solution is a) {x x = π π + kπ x = + kπ, k is an integer} b) To find eight solutions, we can pick k = 1, 0, 1, k = 1 k = 0 π π π + ( 1)π = π ( 1)π = π π π π 0.5 π π π (0)π = π π + (0)π = π y = y = cos(x)
2 100 k = 1 k = π 1π π + (1)π = + ()π = π Ex. sin(θ) = + (1)π = π So, { π, π, π, π solutions to the equation. π + ()π =, 1π, π,, } are eight In the interval [0, π), the sine function is negative in the third and fourth quadrants. In the first quadrant, θ = sin 1 ( ) = π. Thus, π is our reference angle. The angle we need in the third quadrant is π + π = and the angle we need in the fourth quadrant is π π =. The sine function is periodic with period π, so these solutions will repeat every π. Thus, our general solution is a) {x x = + kπ x = + kπ, k is an integer} b) To find eight solutions, we can pick k = 1, 0, 1, k = 1 k = 0 π + ( 1)π = + (0)π = + ( 1)π = π k = 1 k = 10π + (1)π = π + (1)π = So, { π, π, solutions to the equation. Ex. 5cot (x) = First, we need to solve f cot(x): 5cot (x) = 5cot (x) = 5 cot (x) = 1 cot(x) = ± 1 + (0)π = + ()π = 1π + ()π = 17π,, 10π, π, 1π, 17π } are eight
3 101 The cotangent function is positive in quadrant I and negative in quadrant II. Since cot(x) = 1 when tan(x) = 1, then x = π 4 is the reference angle. In quadrant II, the angle we need is π π 4 = π 4. Since the cotangent function is periodic with period of π, then the solutions will repeat every multiple of π. a) {x x = π π + nπ x = nπ, n is an integer}. Since the consecutive angles differ by π, we can write the general solution as: {x x = π 4 + kπ, k is an integer}. b) Eight solutions are { π 4, π 4, π 4, π 4, 4, 7π 4, 9π 4, π 4 }. When solving a trigonometric equation where the argument is a multiple of the variable, we will need to use the general solution f argument and then solve f variable. We will then need to run through values of k until we find all the angles within a specified interval. Solve the following f angles in [0, π). Ex. 4 cos(4θ) 1 = 0 First, solve f cos(4θ): cos(4θ) 1 = 0 cos(4θ) = 1 cos(4θ) = 1 (let x = 4θ) Thus, cos(x) = 1 when x = π x = π π =. Since the period of cosine is π, then the general solution will have the fm of: x = 4 + kπ + kπ Now, solve f θ by dividing by four: 1 + kπ 1 + kπ We need to find all the angles in [0, π). k = π = π = π No 1 k = 0 k = π = π π = 7π π = π = π 1
4 10 k = k = k = π = 1π π = 19π = π = 17π π = π = 9π 1 The solution is { π 1, 1, 7π 1, π 1, 1π 1, 17π 1, 19π 1, π 1 } No Ex. 5 4sin(θ) + = We first need to solve f sin(θ): 4sin(θ) + = 4sin(θ) = 0 sin(θ) = 0 (let x = θ) Thus, sin(x) = 0 when x = 0 π. Since the period of sine is π, then the general solution will have the fm of: x = θ = 0 + kπ π + kπ Since consecutive angles differ by π, then we can state the general solution as: x = kπ Now, solve f θ by dividing by : θ = kπ We need to find all the angles in [0, π). k = 1 θ = 1π = π 0π No k = 0 θ = = 0 k = 1 θ = 1π = π π k = θ = k = k = 5 θ = The solution is {0, π, π Ex. tan(θ + π ) = = π k = 4 θ = k =, π,, }. = π No Let x = θ + π. Since tan(x) = when x = π, then π is our reference angle. The tangent function is negative in quadrant II, so the angle is π π = π. Since the period of tangent is π, then the general solution will have the fm of:
5 x = θ + π = π + kπ Now, solve f θ: θ + π = π + kπ π + kπ + kπ kπ We need to find all the angles in [0, π). k = π = 1 No k = π = π 1 k = π = 7π 1 k = 1 + π = 1π 1 k = 1 + π = 19π 1 k = = 1 No The solution is { π 1, 7π 1, 1π 1, 19π 1 }. Objective : Solving a Trigonometric Equation with a Calculat. Solve the following f angles in [0, π). Ex. 7 cos(x) = 0. Here, we will need to use our calculat to find the value f x: cos(x) = 0. x = cos 1 (0.) = This angle is in the first quadrant. However, the cosine is also positive in the fourth quadrant. Since θ R 0.97, then the angle in the fourth quadrant is π Thus, the solution is { 0.97, 5.559} Ex. 8 tan(x) = Here, we will need to use our calculat to find the value f x: tan(x) = which means x = tan 1 ( ) = This angle is not in [0, π). In that interval, the tangent function is negative in quadrant II, and IV. Since the inverse tangent function gave us the opposite of the reference angle, then θ R = Thus, in the second quadrant, the angle is π
6 104 Ex. 9 = Similarly, in quadrant IV, the angle is π = Hence, the solution is {.044, 5.170}. Due to bad weather, a plane in a holding pattern around the Dallas airpt. The distance d in miles the plane is from the airpt at time t minutes is given by d(t) = 80sin(0.55t) a) When the plane enters the holding pattern, t = 0, how far is it from the airpt? b) During the first 0 minutes after the plane enters the holding pattern, what time(s) t will the plane be exactly 80 miles from the airpt? a) d(0) = 80sin(0.55(0)) + 10 = 80sin(0) + 10 = = 10 The plane was 10 miles from the airpt. b) Set d(t) = 80 and solve: 80sin(0.55t) + 10 = 80 80sin(0.55t) = 50 sin(0.55t) = t = sin 1 ( 0.5) = Thus, the reference angle is The sine is negative in quadrant III and IV, so the angles are π = and π = Thus, 0.55t = kπ kπ, k is an integer. Solving f t yields: t = kπ If k = 0, then t = (0)π kπ (0)π t.995 minutes minutes If k = 1, then t = (1)π (1)π t = t = t 18.5 minutes 1.04 minutes The three times that are within the first twenty minutes are {.995 minutes, minutes, 18.5 minutes}
7 105 Objective : Solving a Trigonometric Equation in Quadratic Fm In solving trigonometric equation in quadratic fm, we will have to use an identity and/ fact befe we can get a series of linear equations involving one trigonometric function to solve. It will also be imptant to make a note of any values that make the iginal equation undefined. Solve f all values in [0, π): Ex. 10 cos (θ) + cos(θ) = 1 cos (θ) + cos(θ) = 1 (get zero on one side) cos (θ) + cos(θ) 1 = 0 Think of x + x 1 = (x 1)(x + 1), so cos (θ) + cos(θ) 1 = (cos(θ) 1)(cos(θ) + 1) Hence, (cos(θ) 1)(cos(θ) + 1) = 0 (solve) cos(θ) 1 = 0 cos(θ) + 1 = 0 cos(θ) = 1 cos(θ) = 1 π π = The solution is { π, π, }. Ex. sin (θ) 8sin(θ) = 0 sin (θ) 8sin(θ) = 0 Think of x 8x = (x )(x + 1), so sin (θ) 8sin(θ) = (sin(θ) )(sin(θ) + 1) Hence, (sin(θ) )(sin(θ) + 1) = 0 (solve) (sin(θ) ) = 0 (sin(θ) + 1) = 0 sin(θ) = No solution The solution is { π }. sin(θ) = 1 Objective 4: Solving Trigonometric Equations Using Identities. Ex. 1 csc (θ) = cot(θ) + 1
8 10 The cosecant and cotangent function is undefined when the sin(θ) = 0 θ = 0 π. Thus, our restrictions are θ 0 π. csc (θ) = cot(θ) + 1 (csc (θ) = cot (θ) + 1) cot (θ) + 1 = cot(θ) + 1 (subtract cot(θ) + 1 from both sides) cot (θ) cot(θ) = 0 (fact cot(θ)) cot(θ)[cot(θ) 1] = 0 (solve) cot(θ) = 0 cot(θ) 1 = 0 cot(θ) = 0 cot(θ) = 1 cot(θ) = 0 when cos(θ) = 0. cot(θ) = 1 when tan(θ) = 1 4 π + π 4 =. None of these values match our 4 π restrictions, so the solution is { π 4, π, 4, π }. Ex. 1 sec(θ) = tan(θ) + cot(θ) The secant and tangent function is undefined when the cos(θ) = 0. The cotangent function is undefined when the sin(θ) = 0 θ = 0 π. Thus, our restrictions are θ 0, π π, π,. sec(θ) = tan(θ) + cot(θ) (write in terms of sine and cosine) 1 cos(θ) = sin(θ) cos(θ) + cos(θ) (multiply by cos(θ)sin(θ)) sin(θ) π 1 sin(θ) cos(θ) cos(θ)sin(θ) = cos(θ)sin(θ) + cos(θ) cos(θ) sin(θ) cos(θ)sin(θ) sin(θ) = sin (θ) + cos (θ) (sin (θ) + cos (θ) = 1) sin(θ) = 1 But, our restrictions say that θ π, so we have to reject our answer. Thus, this equation has no solution.
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