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Unformatted text preview: Math 2312 Homework 8 Section 5.3 1. Find all solutions of the equation in the interval : ) 2 , [ 1 cot 3 2 x Solution: 2 3 2 1 cot ) 4 ( 2 3 2 1 cot ) 3 ( 2 3 2 1 cot ) 2 ( 2 3 2 1 cot ) 1 ( 2 3 2 1 2 3 2 1 3 1 cot 2 3 2 1 2 3 2 1 3 1 cot 3 1 cot 3 1 cot 3 1 cot 1 cot 3 2 2 x x x x x x x x x x In the last line, we have shown that cot x can be equivalent to four different expressions, each of whose numerator is cos x and whose denominator is sin x, since x x x sin cos cot . In equation (1) , 2 1 cos x and 2 3 sin x , which means the angle 3 x . In equation (2), 2 1 cos x and 2 3 sin x , which means the angle 3 4 x . In equation (3), 2 1 cos x and 2 3 sin x , which means the angle 3 2 x . In equation (4), 2 1 cos x and 2 3 sin x , which means the angle 3 5 x . Therefore, all solutions on ) 2 , [ are 3 x , 3 2 x , 3 4 x , and 3 5 x . 2. Find all solutions of the equation in the interval : ) 2 , [ 2 cos sin 2 2 x x Solution: Applying the Pythagorean identity will allow us to factor the left side. 2 1 cos 1 cos 2 cos ) 1 cos 2 ( cos cos cos 2 2 cos cos 2 2 2 cos ) cos 1 ( 2 2 cos sin 2 2 2 2 2 x x x x x x x x x x x x x All angles x on ) 2 , [ such that cos x are 2 x and 2 3 x All angles x on ) 2 , [ such that 2 1 cos x are 3 2 x and 3 4 x Therefore all solutions are: 2 x , 2 3 x , 3 2 x , and 3 4 x 3. Solve the equation. 2 3 2 sin x Solution: Let 2 x , so that 2 3 sin One particular angle where 2 3 sin is 3 4 . Another angle where 2 3 sin is 3 5 . Note that there was no domain restriction for this problem, so we need to include all solutions for x not just those on ) 2 , [ Therefore all angles equivalent to 3 4 are n 2 3 4 where n is an integer. Likewise, all angle equivalent to 3 5 are n 2 3 5 for any integer n....
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This note was uploaded on 10/07/2010 for the course MATH 2312 taught by Professor Garrett during the Summer '10 term at Richland Community College.
 Summer '10
 Garrett
 Math, Calculus

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