Trigonometry, Part 6

Remember what the term argument means? The argument of sin (2x + 7) is 2x + 7, that is an argument of function is the thing the function is operating on. In this case, the sine function is operating on 2x + 7.

So when the argument of a trig function is other than x, the basic method of solving an equation with this function is basically the same as what we did in the last post but an extra bit of algebra is required.

For example, solve 2sin(2x +πœ‹) -1 = 0 for 0 ≀ x ≀ 2πœ‹. The argument of the sine is 2x + πœ‹. We want to first, solve the equation just as we did before, but in terms of the argument 2x + πœ‹:

2sin(2x +πœ‹) -1 = 0 ⟹ sin(2x +πœ‹) = 1/2

From our table of common values, we see that the angle πœ‹/6 has a sine of 1/2. So the basic starting point to find all solutions is 2x +πœ‹ = πœ‹/6. To correctly limit the number of solutions, let’s convert the given domain 0 ≀ x ≀ 2πœ‹ to be in terms of 2x +πœ‹. To do this, algebraically change the inequality so that the middle expression is 2x +πœ‹:

0 ≀ x ≀ 2πœ‹ ⟹ 0 ≀ 2x ≀ 4πœ‹ ⟹ πœ‹ ≀ 2x + πœ‹ ≀ 5πœ‹

So we need to find all the angles between πœ‹ and 5πœ‹ that have a sine of 1/2. Again, a unit circle diagram will help. I will start with πœ‹/6, but realise that πœ‹/6 is not in the required modified domain:

So starting with the basic angle of πœ‹/6, I add multiples of 2πœ‹ until we are outside the given domain. For πœ‹/6, adding 2πœ‹ gives 13πœ‹/6. Adding 2πœ‹ again gives 25πœ‹/6. Adding another 2πœ‹ gives 37πœ‹/6, but 37πœ‹/6 is greater than 5πœ‹ so it is not a valid solution.

So far, the intermediate solutions are 13πœ‹/6 and 25πœ‹/6. The other basic angle found from the unit circle is 5πœ‹/6. Again, this angle is not in our modified domain so we need to add multiples of 2πœ‹ to find the angles within πœ‹ and 5πœ‹. So just like we did for πœ‹/6, adding 2πœ‹ to 5πœ‹/6 gives 17πœ‹/6. Adding another 2πœ‹ gives 29πœ‹/6. Adding another 2πœ‹ puts us outside the domain.

So we have four intermediate solutions:

2x + πœ‹ = 13πœ‹/6
2x + πœ‹ = 25πœ‹/6
2x + πœ‹ = 17πœ‹/6
2x + πœ‹ = 29πœ‹/6

To find the final solutions, we need to solve each of these equations for x:

2x + πœ‹ = 13πœ‹/6 ⟹ x = 7πœ‹/12
2x + πœ‹ = 25πœ‹/6 ⟹ x = 19πœ‹/12
2x + πœ‹ = 17πœ‹/6 ⟹ x = 11πœ‹/12
2x + πœ‹ = 29πœ‹/6 ⟹ x = 23πœ‹/12

A generic method for solving many trig equations is:

  1. Find the basic angles that solve the equation in terms of the argument of the trig function in the equation.
  2. Modify the given domain to be in terms of the argument.
  3. Add or subtract multiples of 2πœ‹ to the basic angles to find all angles within the modified domain. Keep in mind that the basic angles themselves may or may not be in the domain.
  4. Set the argument equal to all solutions and solve for the variable used in the problem (usually πœƒ or x)

There is another way to analyse trig equations which I will show on my next post.