Module 5: Lesson 4

 

Self-Check 4
  1. Questions 1.a., 1.d., 2.c., 3.c., 6, 8, and 10 on pages 320 and 321 of the textbook
    1.  




      1. Therefore, the solutions are




      1. Therefore, the solutions are




      1. Therefore, the solutions are x = 90°, 270°.

        This solution can be verified graphically as shown.

         
        The graph of y equals the tangent of x times the cosine of x times the sine of x minus 1 is shown. It intersects with the x-axis at x equals 90 degrees and x equals 270 degrees. This verifies the solutions found algebraically.

        Because the intersections with the x-axis are the same as the solutions (x = 90°, 270°), the solution can be said to be verified graphically.
    3.  
      1. The equation 3 csc x − sin x = 2 can be converted into a solvable equation in terms of only sine as shown.

         


        Solving the equation gives

         


         


        Therefore, the solution is
    1. Sanesh’s error occurred in the first step. By dividing both sides by cos x, Sanesh failed to find all of the possible solutions. A correct method to solve this equation is shown.

       


       


      Therefore, the solutions, when written in general form, are

       
      x = 30° + 360°n, n ∈ I, x ∈ R
       
      x = 90° + 360°n, n ∈ I, x ∈ R
       
      x = 270° + 360°n, n ∈ I, x ∈ R
       
      x = 330° + 360°n, n ∈ I, x ∈ R
    1. By the zero product property, 7 sin x + 2 = 0 or 3 cos x + 3 = 0 or tan2 x − 2 = 0. Consider the solutions to each of these separately.

      For 7 sin x + 2 = 0,

       


      This will give two solutions for the given domain.

      For 3 cos x + 3 = 0,

       


      This will give one solution for the given domain.

      For tan2 x − 2 = 0,

       


      This will give four solutions, two for  and two for  in the given domain.

      By inspection, all of the solutions are different so there are seven solutions over the interval 0° < x ≤ 360°.


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