Find the values of x, where 0 < x < 360, such that x solves the equation: 8(tan[x])^2 – 5(sec[x])^2 = 7 + 4sec[x]
This question tests a variety of mathematical techniques and knowledge which is integral to A-level mathematics.
Note that the highest power of a variable function in the mentioned equation, is 2. It is therefore quite likely that we will have to solve a quadratic equation to find x. We know how to solve quadratic equations from GCSE. Therefore, if we can manipulate the equation such that we have only one trigonometric variable function, then we will be able to solve the problem.
We first notice that tan[x] = sin[x]/cos[x] and that sec[x] = 1/cos[x]
Hence the equation can be read as:
8(sin[x]/cos[x])2 - 5(1/cos[x])2 = 7 + 4(1/cos[x])
Solving a quadratic equation with fractions is always going to be a messy business. We notice that cos[x] is the denominator of several fractions in the equation. To get rid of the denominators, we times through by (cos[x])2:
8(sin[x])2 - 5 = 7(cos[x])2 + 4cos[x] {*}
This already looks easier to solve than the initial equation. If we could ensure that the only variable function was sin[x] or cos[x] then we could solve the equation as a quadratic! Recall that:
(sin[x])2 + (cos[x])2 =1
Therefore:
(sin[x])2 = 1 - (cos[x])2
Substituting this value into {*}, we find that:
8(1 - (cos[x])2) - 5 = 7(cos[x])2 + 4cos[x]
By rearranging appropriately, we get the quadratic equation:
15(cos[x])2 + 4cos[x] - 3 = 0
Just as in GCSE, we find that this is equivalent to:
(3cos[x] - 1)(5cos[x] + 3) = 0
so
(1) cos[x] = 1/3 or (2) cos[x] = -3/5
We consider cases (1) and (2) separately.
Case 1: cos[x] = 1/3
Using a calculator we see that x = cos-1[1/3] = 70.5 (to 1 decimal place). However we want every value of x such that 0 < x < 360.
Note that if cos[x] = y, then for any integer k:
x = 360k ± cos-1[y]
This is the general solution for cosine.
By using the general solution for cosine we see that
x = 360 - cos-1[1/3] = 289.5 (to 1 decimal place)
is also a solution.
Case 2: cos[x] = -3/5
Using a calculator we see that x = cos-1[-3/5] = 126.9 (to 1 decimal place).
Using the general formula for cosine we see that
x = 360 - cos-1[-3/5] = 233.1 (to 1 decimal place).
is also a solution in the interva 0 < x < 360.
Hence we see that the complete set of values of x which solve 8(tan[x])2 – 5(sec[x])2 = 7 + 4sec[x] in the range 0 < x < 360 is:
x = 70.5, 126.9, 233.1, 289.5
Note: In the answer we used the general solution for cosine. This may have to be justified. This can be explained intuitively with a graph of the cosine function. Graphs and diagrams are often extremely helpful in understanding and answering questions.
