Explanation Given: The function y 1 = sin − 1 x − cos − 1 x and y 2 = π 6 has a zero of point 3 2 ≈ 0.86602 . Graph: Consider the functions y 1 = sin − 1 x − cos − 1 x and y 2 = π 6 , which have a zero of point 3 2 ≈ 0.86602 . Now, y 1 = y 2 sin − 1 x − cos − 1 x = π 6 x = sin ( cos − 1 x + π 6 ) Let cos − 1 x = u . So, x = sin ( u + π 6 ) = sin u cos π 6 + cos u sin π 6 = 3 2 sin u + 1 2 cos u The range of sine function over the interval [ − π 2 , π 2 ] is − π 2 ≤ sin x ≤ π 2 = − π 2 ≤ cos − 1 x + π 6 ≤ π 2 = − π 2 − π 6 ≤ cos − 1 x ≤ π 2 − π 6 = − 2 π 3 ≤ cos − 1 x ≤ π 3 Thus, for both 0 ≤ cos − 1 x ≤ π and − 2 π 3 ≤ cos − 1 x ≤ π 3 , the intersection yields 0 ≤ cos − 1 x ≤ π 3

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- for Trigonometry

11th Edition

ISBN: 9780135909140

Author: Lial, Margaret, HORNSBY, John, SCHNEIDER, David, DANIELS, Callie

Publisher: PEARSON

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Chapter 6.4, Problem 46E

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To graph: The solution of the provided expression by use of graphing calculator if The function y1=sin−1x−cos−1x and y2=π6 has a zero of point 32≈0.86602.

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Chapter 6.4, Problem 45E

Chapter 6.4, Problem 47E

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MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- for Trigonometry

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To graph: The solution of the provided expression by use of graphing calculator if The function y 1 = sin − 1 x − cos − 1 x and y 2 = π 6 has a zero of point 3 2 ≈ 0.86602 .

Solution Summary: The author analyzes the solution of the provided expression by use of graphing calculator if the functions y_1=mathrmsin-1x-

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To graph: The solution of the provided expression by use of graphing calculator if The function y1=sin−1x−cos−1x and y2=π6 has a zero of point 32≈0.86602.

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Chapter 6 Solutions