Module 5: Lesson 2

 

Self-Check 2
  1. “Your Turn” from “Example 2” on page 270 of the textbook

    An algebraic solution is shown.

     


    Since the sine function is positive in quadrants 1 and 2, a second possible value of x can be calculated.

     


    Because the b value is the period is 8. Thus, the general solution for the equation is x ≈ 0.432 + 8n, n ∈ I, x ∈ R and x ≈ 3.567 + 8n, n ∈ I, x ∈ R.

    A graphical solution is shown.

    The graph of y equals 6 times the sine of pi divided by 4 times x all plus 8 is shown as the linear graph of y equals 10. The points of intersection at (0.432, 10) and (3.567, 10) are also labelled.

    Because the b value is the period is 8. Thus, the general solution for the equation is x ≈ 0.432 + 8n, n ∈ I, x ∈ R and x ≈ 3.567 + 8n, n ∈ I, x ∈ R.



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