Stitz-Zeager_College_Algebra_e-book

# Hence arcsin 2 56 arcsin 2 radians 10819 6 3 3 6 an

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Unformatted text preview: and check your answers analytically. List the solutions which lie in the interval [0, 2π ) and verify them using a graphing utility. √ 1. cos(2x) = − 2. csc 1 3x 3 2 −π = x 2 3. cot (3x) = 0 √ 2 5. tan = −3 4. sec2 (x) = 4 6. sin(2x) = 0.87 Solution. √ 1. On the interval [0, 2π ), there are two values with cosine − √ 3 2, namely 3 2 5π 6 and 7π 6. Hence, we begin solving cos(2x) = − by setting the argument 2x equal to these values and add π π multiples of 2π (the period of cosine) which yields 2x = 56 +2πk or 2x = 76 +2πk for integers k . Solving for x gives x = 5π + πk or x = 7π + πk for integers k . To check these answers 12 12 analytically, we substitute them into the original equation. For any integer k we have 1 2 If c is isn’t in the range of T , the equation has no real solutions. See the comments at the beginning of Section 10.5 for a review of this concept. 730 Foundations of Trigonometry cos 2 5π 12 + πk = cos = cos √ =− 5π 6 5π 6 + 2πk (he period of cosine is 2π .) 3 2 √ π π Similarly, we ﬁnd cos 2 7π + πk = cos 76 + 2πk = cos 76 = − 23 . To determine which 12 of our so...
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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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