Stitz-Zeager_College_Algebra_e-book

This means if a h we are always guaranteed to have at

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Unformatted text preview: + 2πk = csc π = 2 and 4 4 4 √ π π π csc 1 21π + 6πk − π = csc 74 + 2πk − π = csc 34 + 2πk = csc 34 = 2. Despite 3 4 having infinitely many solutions, we find that none of them lie in [0, 2π ). To verify this graphically, we use a reciprocal identity to rewrite the cosecant as a sine and we find that √ y = sin 11x−π and y = 2 do not intersect over the interval [0, 2π ). (3 ) √ y = cos(2x) and y = − 3 2 y= 1 sin( 1 x−π ) 3 and y = √ 2 3. In the interval (0, π ), only one value, π , has a cotangent of 0. Since the period of cotangent is π , 2 the solutions to cot(3x) = 0 are 3x = π + πk for integers k . Solving for x yields x = π + π k . 2 6 3 Checking our answers, we have that for any integer k , cot 3 π + π k = cot π + πk = 6 3 2 cot π = 0. As k runs through the integers, we obtain six answers, corresponding to k = 0 2 π π π through k = 5, which lie in [0, 2π ): x = π , π , 56 , 76 , 32 and 11π . To confirm these graphically, 62 6 we must be careful. On many calculators, there is no function button for...
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