Solution to Exercise 7.1-2

The problem was to solve

   tan(x)  =  A cos(x)

for x, where -π/2 ≤ x ≤ π/2 and A > 0.

Step 1: Multiply out the denominator. Observe that tan(x) = sin(x)/cos(x). So if you multiply through by cos(x) you get the more tractable equation of

   sin(x)  =  A cos²(x)

Step 2: Apply an identity to cos²(x). Since sin²(x) + cos²(x) = 1, you can replace cos²(x) with 1 - sin²(x). So you have

   sin(x)  =  A - A sin²(x)

Step 3: Substitute for sin(x). What you've got here is a quadratic in sin(x). Make it easy on yourself by substituting u = sin(x). You can substitute back later. Now you have

   u  =  A - Au²

or equivalently

   Au² + u - A  =  0

Step 4: Apply the quadratic formula. Use it to solve for u.

               _______
         -1 ± √1 + 4A²
   u  =               
               2A

Step 5: Decide which root makes sense. As you can see, the quadratic above has two solutions. Which one is the one we are looking for? Remember that u = sin(x), and -1 ≤ sin(x) ≤ 1 for all x. Hence -1 ≤ u ≤ 1. Observe that when you do some algebraic massaging on the quadratic solutions, you get

         -1
   u  =     ±
         2A

 1
    + 1
4A²

The expression inside the square root is already greater than 1. Hence the square root itself is also greater than 1. If A is positive, as the problem stipulates, then subtracting the square root from a negative value will result in something less than -1 (hence its magnitude is greater than 1, and you can't take the arcsine of it). So we must conclude that we have to use the positive square root in this case. So

         -1
   u  =     +
         2A

 1
    + 1
4A²

Observe that if we had allowed A to be negative, we would have had to use the other root for that case.

Can you prove that the above expression for u will always be in the open interval of 0 to 1 no matter what positive value you choose for A?

Step 6: Use the inverse sine to solve for x. Since we substituted earlier u = sin(x), we have

   x  =  sin^-1(u)

Observe that the principle value of the inverse sine gives us x in the range that the problem asked for.

Step 7: Substitute the solution you got for u. So you have


   x  =  sin^-1

 -1
    +
 2A

 1
    + 1
4A²

which completes the solution.

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