Stitz-Zeager_College_Algebra_e-book

Think about this before reading further example 1123

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Unformatted text preview: 2)2 = 4 for all integers k . The same holds for the family x = 23 + πk . The 3 π π solutions which lie in [0, 2π ) come from the values k = 0 and k = 1, namely x = π , 23 , 43 3 5π and 3 . To confirm graphically, we use a reciprocal identity to rewrite the secant as cosine. 1 The x-coordinates of the intersection points of y = (cos(x))2 and y = 4 verify our answers. y= cos(3x) sin(3x) and y = 0 y= 1 cos2 (x) and y = 4 5. Our first step in solving tan x = −3 is to consider values in the interval − π , π with a 2 22 tangent of −3. Since −3 isn’t among the ‘common values’ for tangent, we need the arctangent function. The period of the tangent function is π , so we get x = arctan(−3)+ πk for integers k . 2 Multiplying through by 2 gives us our solution x = 2 arctan(−3)+2πk for integers k . To check − our answer, we note that for any integer k , tan 2 arctan(2 3)+2πk = tan (arctan(−3) + πk ) = tan (arctan(−3)) = −3. To determine which of our answers lie in the interval [0, 2π ), we begin substituting integers k into the expression x = 2 arctan(−3) + 2πk . When k = 0, we get x = 2 ar...
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