Karl's Calculus Tutor - Solution to Exercise 7.1-2

Solution to Exercise 7.1-2

The problem was to solve

   tan(x)  =  A cos(x)

for x, where -p/2 £ x £ p/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 ± Ö1 + 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 adding a positive value to something that is already greater than 1 guarantees a sum greater than 1. So we must conclude that we have to use the negative 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 -1 to 0 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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