Question

Solve.

20sec^2x− 3secx−9=0
on [0,2π)

x=π/4, π/2, 4π/3, and 5π/3
x=π/2
x=π/4 and 5π6
x=π/4
There are no solutions.

Answer 1

To solve the equation \( 20 \sec^2 x - 3 \sec x - 9 = 0 \), we will make a substitution for \(\sec x\). Let \( y = \sec x \). The equation becomes:

\[
20y^2 - 3y - 9 = 0
\]

Now, we can use the quadratic formula to solve for \( y \):

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 20 \), \( b = -3 \), and \( c = -9 \). Calculating the discriminant:

\[
b^2 - 4ac = (-3)^2 - 4 \cdot 20 \cdot (-9) = 9 + 720 = 729
\]

Now applying the quadratic formula:

\[
y = \frac{-(-3) \pm \sqrt{729}}{2 \cdot 20} = \frac{3 \pm 27}{40}
\]

Calculating the two possible values for \( y \):

1. \( y_1 = \frac{30}{40} = \frac{3}{4} \)
2. \( y_2 = \frac{-24}{40} = -\frac{3}{5} \)

Since \( y = \sec x = \frac{1}{\cos x} \), we only consider positive values because the secant function is undefined for negative inputs in the context of real angles.

The valid solution is:

\[
y = \frac{3}{4} \Rightarrow \sec x = \frac{3}{4}
\]

Thus, we can find \( \cos x \):

\[
\cos x = \frac{4}{3}
\]

However, since \(\cos x\) must be in the range \([-1, 1]\), \(\sec x = \frac{3}{4}\) leads to an invalid cosine value.

Now we consider the other value we calculated:

The case \( y = -\frac{3}{5} \) is also not possible, as \( \sec x \) must always be positive for real angles.

Thus, since both potential secant values lead to invalid cosine ranges, we can conclude that:

There are no solutions in the interval \( [0, 2\pi) \).

Therefore, the answer is:

**There are no solutions.**