### Video Transcript

Find the set of values satisfying
the cos of 𝜃 minus 105 is equal to negative a half, where 𝜃 is greater than zero
degrees and less than 360 degrees.

In order to find the solutions to a
trig equation in a given interval, we begin by finding a particular solution. In this case, the table of exact
trigonometric values can help. We will first redefine the argument
of the function by letting 𝛼 equal 𝜃 minus 105 such that the cos of 𝛼 is equal to
negative one-half and 𝜃 is equal to 𝛼 plus 105. We can then amend the interval over
which our solutions are valid by adding 105 to each part of the inequality; 𝛼 is
greater than 105 degrees and less than 465 degrees. Filling in the table for the exact
values of cos 𝛼, we can see that cos 𝛼 equals a half when 𝛼 is 60 degrees. However, there are no values of 𝛼
in the table such that cos 𝛼 is equal to negative a half.

By sketching the graph of the
cosine function along with the lines 𝑦 equals one-half and 𝑦 equals negative a
half, we can find the associated value of 𝛼. It appears on the graph that there
might be three values between 105 and 465 degrees. As the graph has rotational
symmetry between zero and 180 degrees about 90 degrees, zero, the first solution is
equal to 180 minus 60. This is equal to 120 degrees, which
lies in the required interval. Next, using symmetry of the curve,
we have 𝛼 is equal to 180 plus 60. This is equal to 240 degrees, which
also lies in the given interval. The third solution corresponds to
120 plus 360 degrees. However, this value of 480 degrees
lies outside of our interval for 𝛼. Hence, the solutions to cos 𝛼
equals negative one-half are 𝛼 equals 120 degrees and 𝛼 equals 240 degrees.

We can now calculate the
corresponding values of 𝜃. 120 plus 105 is equal to 225, and
240 plus 105 is 345. The set of values that satisfy cos
of 𝜃 minus 105 equals negative one-half are 225 degrees and 345 degrees. An alternative technique to find
the particular solution to cos of 𝛼 equals negative one-half is to use the inverse
cosine function such that 𝛼 is equal to the inverse cos of negative one-half, which
is equal to 120 degrees. From this point, we would use the
same steps to find the other solutions. This could also have been done
using the unit circle, which would lead us to the general rule: the cos of 𝜃 is
equal to the cos of 360 degrees minus 𝜃.

Using the symmetry of the unit
circle and periodicity of the cosine function, we can quote formulas for the general
solution to equations involving this function. In the same way as we have already
seen for the sine function, then the set of all solutions to cos 𝜃 equals 𝐶 is 𝜃
is equal to 𝜃 sub one plus 360𝑛 and 𝜃 is equal to 360 minus 𝜃 sub one plus 360𝑛
for all integer values of 𝑛. Once again, if 𝜃 is measured in
radians, we replace 360 degrees with two 𝜋 radians.