19 a3 30 4181 30 c4 a3 13819 b 570 b

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Unformatted text preview: ine or arcsine is needed, consider using the ‘reference angle’ approach as demonstrated in Example 10.6.7 numbers 1 and 3. • If T (x) = sec(x) or T (x) = csc(x), convert to cosine or sine, respectively, and solve as above. • If T (x) = tan(x), solve for x on − π , π , using the arctangent as needed, and add integer 22 multiples of the period π . • If T (x) = cot(x), solve for x on (0, π ), using the arccotangent as needed, and add integer multiples of the period π . Using the above guidelines, we can comfortably solve sin(x) = 1 and find the solution x = π + 2πk 2 6 π 1 or x = 56 + 2πk for integers k . How do we solve something like sin(3x) = 2 ? One approach is to solve the equation for the argument 2 of the sine function, in this case 3x, to get 3x = π + 2πk or 6 π 3x = 56 + 2πk for integers k . To solve for x, we divide the expression through by 3 and obtain π π π x = 18 + 23 k or x = 5π + 23 k for integers k . This is the technique employed in the example below. 18 Example 10.7.1. Solve the following equations...
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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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