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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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