EXAMPLE 3 Find, correct to five decimal places, the positive root of the equation 4cos(x) = x. SOLUTION We first rewrite the equation in standard form: 4cos(x) - x = 0 Therefore, we let f(x) = 4cos(x) - x. Then, f '(x) = so Newton's method becomes 4 cos (x) – x 4 sin (r) +1 4 cos (x) – x In+1 = I, - -4sin (z) -T = Ľ,+ Video Example ) In order to guess a suitable value for x1, we sketch the graphs of y = 4cos(x) and y = x in the figure. It appears that they intersect at a Tutorial point whose x-coordinate is near 1, so let's take x, - 1 as a convenient first approximation. Then remembering to put our calculator in Online Textbook radian mode, we get X2 1.26597 X3 1.25238 X4 1.25235 Xg 1.25235 Since the last two agree to five decimal places, we conclude that the root of the equation, correct to five decimal places, is

EXAMPLE 3 Find, correct to five decimal places, the positive root of the equation 4cos(x) = x. SOLUTION We first rewrite the equation in standard form: 4cos(x) - x = 0 Therefore, we let f(x) = 4cos(x) - x. Then, f '(x) = so Newton's method becomes 4 cos (x) – x 4 sin (r) +1 4 cos (x) – x In+1 = I, - -4sin (z) -T = Ľ,+ Video Example ) In order to guess a suitable value for x1, we sketch the graphs of y = 4cos(x) and y = x in the figure. It appears that they intersect at a Tutorial point whose x-coordinate is near 1, so let's take x, - 1 as a convenient first approximation. Then remembering to put our calculator in Online Textbook radian mode, we get X2 1.26597 X3 1.25238 X4 1.25235 Xg 1.25235 Since the last two agree to five decimal places, we conclude that the root of the equation, correct to five decimal places, is

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.2: Trigonometric Equations

Problem 84E

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