Explanation Given: s i n 2 x + c o s x = 0 Also, value of x is such that 0 ≤ x ≤ 2 π Formula: s i n 2 x = 2 s i n x c o s x Calculation: The equation is given as s i n 2 x + c o s x = 0 Putting the value of s i n 2 x in above equation, we get 2 s i n x c o s x + c o s x = 0 Taking c o s x common, we get c o s x ( 2 s i n

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 1, Problem 51CRE

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The solution of equation sin 2x + cos x = 0 for the value of x such that 0 ≤ x ≤ 2π

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Chapter 1, Problem 50CRE

Chapter 1, Problem 52CRE

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(b) Find the (instantaneous) rate of change of y at x = 5. In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the following limit. lim h→0 - f(x + h) − f(x) h The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule defining f. f(x + h) = (x + h)² - 5(x+ h) = 2xh+h2_ x² + 2xh + h² 5✔ - 5 )x - 5h Step 4 - The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x). - f(x + h) f(x) = = (x² x² + 2xh + h² - ])- = 2x + h² - 5h ])x-5h) - (x² - 5x) = ]) (2x + h - 5) Macbook Pro

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Chapter 1 Solutions

CALCULUS (CLOTH)

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The solution of equation s i n 2 x + c o s x = 0 for the value of x such that 0 ≤ x ≤ 2 π

Solution Summary: The author explains how to determine the value of x, based on the equation.

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The solution of equation sin 2x + cos x = 0 for the value of x such that 0 ≤ x ≤ 2π

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Chapter 1 Solutions