Math 150 T16-Trigonometric Equations Page 1 MATH 150 TOPIC 16 TRIGONOMETRIC EQUATIONS In calculus, you will often have to find the zeros or x-intercepts of a function. That is, you will have to determine the values of x that satisfy the equation f(x) = 0. This section reviews some techniques for solving f(x) = 0 when f(x) is a trigonometric function. A good grasp of Review Topics 1, 1, 14, and 15 will prove to be very useful. Also, remember that after you solve an equation you can always check your answer. We first consider the following equation. Solve for x: sinx = 1, 0 x π. The graph of sin x for 0 x π and the line y = 1 appear below. 1 f(x) =sinx 1 π y = 1 π 1 1 Fig. 16.1. When 0 x π, we see that there are two values of x where sin x = 1 ; namely, x = π 6 and x = 5π 1. Similarly, the equation sin x = has 6 solutions x = 5π 4 and x = 7π. The equation sin x = has no solutions 4 since the range of sin x =[ 1, 1] (i.e., the outputs must be between 1 and 1).
Math 150 T16-Trigonometric Equations Page A slightly more difficult equation to solve is the following. Solve for x: sinx =.4, 0 x π. In the previous equations we were able to solve for x exactly. Now we must use a calculator, which yields that x = arcsin(.4) =.41 radians. The angle.41 radians is in the first quadrant. Fig. 16.1 implies there is a second value of x where sin x =.4. Using the concept of the reference angle (Review Topic 9a), this second value lies in the second quadrant and = (π.41) or.7 radians. 16.1. Solve the following equations where 0 x π. a) cos x = 1 b) sin x =0 c) cos x =.65 (use a calculator) Next, consider a problem of the form: sin x = 1, 0 x π. Although we plug in x, the sine sees x. In other words, if 0 x π, the sine sees 0 x 4π. On this extended interval [0, 4π], sin( ) = 1 at π 6, 5π 6, π 6 +π, 5π 6 +π. This means x = π 6, 5π 6, π 6 +π, 5π 6 +π. Thus, x = π 1, 5π 1, π 1π + π or 1 1, 5π 17π + π or 1 1. 16.. Solve the following equation. cos x =1, 0 x π.
Math 150 T16-Trigonometric Equations Page Many times trigonometric equations will be quadratic in form. Consider the examples below. cos x = cos x 1 versus r =r 1 sin x sin x =0 versus r r =0 This suggests solving the trigonometric equation as if it were a quadratic equation. Let s do the first example. Solve: cos x = cos x 1, 0 x x. cos x cosx + 1 = 0 Get 0 on one side ( cos x 1)(cos x 1) = 0 Factor cosx 1=0 cos x 1 = 0 Set each factor = 0 cosx =1 cos x =1 cos x = 1 x =0, π x = π, 5π Solutions: x =0, π, 5π,π. 16.. Solve the following equations. a) sin x 4 =0, 0 x π b) tan x + tan x =0, π <x<π. Trigonometric identities can be involved in solving trigonometric equations. Consider the following problem. Solve: cos x +cosx =0, 0 x π.
Math 150 T16-Trigonometric Equations Page 4 The fact that one function has argument x and the other function has argument x suggests that we try to get the same argument in both functions. Using the identity cos x = cos x 1, we proceed as follows. cos x +cosx =0, 0 x π cos x +cosx 1=0 ( cos x 1)(cos x +1)=0 cosx 1=0 cos x +1=0 cos x = 1 cos x = 1 x = π, 5π x = π Solutions: x = π, π, 5π. Question: Why did we choose the identity cos x = cos x 1inthe preceding problem? There are other ways to write cos x. Why wouldn t they work? As another example, consider sin x cos x = 1, 0 x π. Since sin x =sinx cos x, wehave sin x = 1, or sin x = 1, 0 x π. Notice that the sine sees x as in an earlier example. This means that when 0 x π, the sine sees 0 x 4π. Thus sin(x) = 1 when x = π or x = 7π, which implies x = π 4 or 7π 4.
Math 150 T16-Trigonometric Equations Page 5 16.4. Solve the following equations. a) 1 + sin x = cos x, 0 x π. b) tan x =sinx, π <x<π. The next example is a bit tricky. Solve: cos x +sinx =0, 0 x π. In previous problems, it was helpful to get 0 on one side and then work with the other side. For this equation, we instead move sin x to the right hand side and obtain cos x = sin x, 0 x π. Now we remember that cos x and sin x are coordinates of a point P as it moves around the unit circle (Review Topic 11). So, the equation cos x = sin x is equivalent to asking when the coordinates of P have the same absolute value but differ in sign. A moment s thought reveals that x must be π 4 or 7π 4. Another solution method is as follows. Solve: cos x = sin x, 0 x π. Divide each side by ( cos x) and obtain 1 = tan x, 0 x π, x π, π. NOTE: We must exclude π and π because tan x is not defined for these values. Also, π or π do not solve the original equation. Thus we have tan x = 1, 0 x π, x π, π.
Math 150 T16-Trigonometric Equations Page 6 Solutions: x = π 4, 7π 4. 16.5. Solve the equation sin x cos x =0, 0 x π. In the previous problems, the equations were solved on some specified interval. However, the interval can be unrestricted. For example, Solve: sin x = 1, for any x. For problems like this, we must remember that the sine function is periodic with period π. So we first solve the equation on the interval 0 x<π, and then we add multiples of π. The solution is: x = π 6 +kπ or 5π +kπ, for any integer k 6 (k can be positive or negative). 16.6. Solve the following equations where x is arbitrary. a) cos x cos x =0. b) tan x 1=0.
Math 150 T16-Trigonometric Equations Page 7 ANSWERS to PRACTICE PROBLEMS (Topic 16 Trigonometric Equations) 16.1. a) x = π, 5π b) x =0,π,π c) Solve cos x =.65, 0 x π. Using arccos( ) on your calculator (in radian mode), you get x =.86 radians. However, cos x is also positive in the fourth quadrant where π x π. The reference angle concept (Review Topic 9a) implies that x =π.86 = 5.4 radians. Solutions: x =.86 or 5.4 16.. Solve cos(x) =1,0 x π. The cosine sees (x), which means 0 (x) 4π. On this interval, cos( ) = 1 when ( ) = 0, π, 4π. Thus,(x) =0, π, or4π, which implies x =0,π,orπ. 16.. a) Method 1. Factor the left hand side and obtain ( )( ) sin x sin x + =0, 0 x π. sin x =0 sin x + =0 sin x = sin x = x = π, π x = 4π, 5π Solutions: x = π, π, 4π, 5π
Math 150 T16-Trigonometric Equations Page 8 Method. Solve for sin x. sin x = 4 sin x = ±. So, sin x = sin x = Now, continue as above. b) Solve tan x + tan x =0, π <x<π. tan x(tan x +1)=0 tan x =0 tan x +1=0 x =0 tan x = 1 x = π 4 Solutions: x = π 4, 0. 16.4. a) Solve 1+sinx = cos x, 0 x π = (1 sin x) = sin x sin x +sinx 1=0 ( sin x 1)(sin x +1)=0 sinx 1=0 sin x +1=0 sin x = 1 sin x = 1 x = π 6, 5π 6 x = π Solutions: x = π 6, 5π 6, π.
Math 150 T16-Trigonometric Equations Page 9 b) Solve tan x =sinx, π <x<π. sin x sinx =0 cos( x ) 1 sin x cos x =0 sin x =0 x =0 1 cos x =0 1 cos x = cos x = 1 x = π, π Solutions: x = π, 0, π. 16.5. Solve sin x cos x =0,0 x π. Method 1: sin x =cosx; 0 x π. Since cos x and sin x are the x and y coordinates of a point P on the unit circle, this question is equivalent to asking when the coordinates have the same value. The answer is when P is in the first and third quadrants where x = π 4 and 5π 4, respectively. Method : Divide by cos x and obtain tan x 1=0, 0 x π, x π, π tan x =1 x = π 4, 5π 4.
Math 150 T16-Trigonometric Equations Page 10 16.6. a) Solve cos x cos x = 0, where x is arbitrary. ( cos x cos x 1 ) =0 cos x =0 cos x 1 =0 x = π +kπ, or cos x = 1 x = π +kπ x = π +kπ, 5π +kπ Equivalently, x = π + kπ Solutions: x = π + kπ, π +kπ, 5π +kπ, for any integer k. b) Solve tan x 1=0. tan x = 1 tan x = ± 1 tan x = 1 tan x = 1 x = π 6 + kπ x = π 6 + kπ Solution: x = ± π + kπ, for any integer k. (Remember that tan x 6 has period π. This means that we only need to add multiples of π, notπ.) Beginning of Topic Skills Assessment