Stitz-Zeager_College_Algebra_e-book

These four cases exemplify all of the possibilities

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Unformatted text preview: lutions lie in [0, 2π ), we substitute integer values for k . The solutions we keep come from the values of k = 0 and k = 1 and are x = 5π , 7π , 17π and 19π . To confirm these 12 12 12 12 √ answers graphically, we plot y = cos(2x) and y = − 23 over [0, 2π ) and examine where these two graphs intersect. We see that the x-coordinates of the intersection points correspond to our exact answers. 2. As we saw in Section 10.3, equations involving cosecant are usually best handled by converting √ √ 1 1 the cosecants to sines. Hence, we rewrite csc 1 x − π = 2 as sin 3 x − π = √2 = 22 . 3 √ π There are two values in [0, 2π ) with sine 22 : π and 34 . Since the period of sine is 2π , we get 4 π 1 3π 1 3 x − π = 4 +2πk or 3 x − π = 4 +2πk for integers k . Solving for x, we get our general solution x = 15π + 6πk or x = 21π + 6πk for integers k . Checking these answers, we get that for any 4 4 √ 1 π integer k , csc 3 15π + 6πk − π = csc 54 + 2πk − π = csc...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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