Video Transcript
Find the set of values satisfying
six cos squared 𝜃 minus seven cos 𝜃 minus five equals zero, where 𝜃 is greater
than or equal to zero and less than or equal to 360 degrees. Give your answers to the nearest
minute.
Notice that we have a trigonometric
equation and that this trigonometric equation looks a little bit like a
quadratic. In fact, we’re going to treat it a
lot like one. We’re going to begin by defining 𝑥
to be cos 𝜃. And replacing cos 𝜃 with 𝑥 in our
original trigonometric equation and we find that six 𝑥 squared minus seven 𝑥 minus
five must be equal to zero. We’re going to begin then by
solving this equation for 𝑥.
And in fact, the expression six 𝑥
squared minus seven 𝑥 minus five is fully factorable. We know it can be written as the
product of two polynomials. The first term in each polynomial
must have a product of six 𝑥 squared, whereas the second term must have a product
of negative five. Then, when we multiply the outer
terms and inner terms and find their sum, we need the solution to be negative seven
𝑥. The only way for these criteria to
be satisfied is if our polynomials are three 𝑥 minus five and two 𝑥 plus one.
Now, the product of these two
polynomials is equal to zero. And the only way for this to be
true is if either three 𝑥 minus five is equal to zero or two 𝑥 plus one is equal
to zero. We solve this first equation by
adding five to both sides and then we divide through by three. And we see that 𝑥 is equal to
five-thirds. Similarly, we solve the second
equation for 𝑥 by subtracting one from both sides and then dividing by two. And we see that 𝑥 here is equal to
negative one-half. Now, we originally said that 𝑥 was
equal to cos 𝜃. So we can now say that cos 𝜃 must
be equal to five-thirds or negative one-half.
We’re going to look at solving each
of these equations for 𝜃. However, we know the absolute
maximum value of cos 𝜃 to be one. So this means there’s no value of
𝜃 such that cos 𝜃 is equal to five-thirds. And here, we say 𝜃 is
undefined. We can solve the second equation
though by finding the inverse cos of both sides. So 𝜃 is equal to the inverse cos
of negative one-half. That gives us that 𝜃 is equal to
120 degrees. And so, we have found one value of
𝜃, which satisfies this equation. However, it’s not the only one. And there’re a couple of different
ways that we can find the second.
One method involves sketching the
graph of 𝑦 equals cos 𝜃. We’re solving the equation cos 𝜃
equals negative one-half. So we add the line 𝑦 equals
negative one-half to our diagram. We saw that the first solution, the
first value of 𝜃 for which this is the case, is 120 degrees. So that’s the first point of
intersection between the graph of cos 𝜃 and negative one-half. The second solution is over
here. Now, in fact, we can use the fact
that the graph is completely symmetrical about the line 𝜃 equals 180 degrees to
find this value. We subtract 120 degrees from
360. And then, we find the other value
of 𝜃 in the interval 𝜃 is greater than or equal to zero and less than or equal to
360 is 240 degrees.
The other method you might be
confident with is the cos diagram. We are solving the equation cos 𝜃
equals negative one-half. So we used our cos diagram to
identify the fact that cos is positive here and here. This, therefore, means it’s
negative here and here. And so, our first solution for 𝜃
lies in the second quadrant. We see that our second solution
lies in the third quadrant. We know the angles around a point
sum to 360 degrees. So we can find our second solution
once again by subtracting 120 from 360 to get 240 degrees.
And so, the set of values
satisfying the equation six cos squared 𝜃 minus seven cos 𝜃 minus five equals
zero, where 𝜃 is greater than or equal to zero and less than or equal to 360
degrees are 120 degrees and 240 degrees.
