Number of real solution of the equation `(tan x+1) (tan x+3) (tan x+5) (tan x+7)=33`

Options

A. will be two in the interval `[-pi//2, pi//2]`
B. will be four in the interval `[-pi//2, pi//2]`
C. will be three in the interval `[-pi//2, pi]`
D. will be four in the interval `[-pi//2, pi]`

Explanation

To solve the equation \((\tan x + 1)(\tan x + 3)(\tan x + 5)(\tan x + 7) = 33\), we can follow these steps:

### Step 1: Substitute \(y = \tan x\)

Let \(y = \tan x\). The equation can be rewritten as:

\[ (y + 1)(y + 3)(y + 5)(y + 7) = 33 \]

### Step 2: Expand the left-hand side

We can expand the left-hand side:

\[ (y + 1)(y + 7) = y^2 + 8y + 7 \]
\[ (y + 3)(y + 5) = y^2 + 8y + 15 \]

Now, multiply these two results:

\[ (y^2 + 8y + 7)(y^2 + 8y + 15) \]

Let \(p = y^2 + 8y\). Then:

\[ (y^2 + 8y + 7)(y^2 + 8y + 15) = (p + 7)(p + 15) = p^2 + 22p + 105 \]

### Step 3: Set the equation equal to 33

Now, we set this equal to 33:

\[ p^2 + 22p + 105 = 33 \]

### Step 4: Rearrange the equation

Rearranging gives:

\[ p^2 + 22p + 72 = 0 \]

### Step 5: Solve the quadratic equation

Now we can solve this quadratic using the quadratic formula:

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

Here, \(a = 1\), \(b = 22\), and \(c = 72\):

\[ p = \frac{-22 \pm \sqrt{22^2 - 4 \cdot 1 \cdot 72}}{2 \cdot 1} \]
\[ p = \frac{-22 \pm \sqrt{484 - 288}}{2} \]
\[ p = \frac{-22 \pm \sqrt{196}}{2} \]
\[ p = \frac{-22 \pm 14}{2} \]

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

1. \(p = \frac{-22 + 14}{2} = \frac{-8}{2} = -4\)
2. \(p = \frac{-22 - 14}{2} = \frac{-36}{2} = -18\)

### Step 6: Determine valid solutions for \(p\)

Since \(p = y^2 + 8y\), we need \(p \geq -16\). Thus, we discard \(p = -18\) and accept \(p = -4\).

### Step 7: Substitute back to find \(y\)

Now substitute back to find \(y\):

\[ y^2 + 8y + 4 = 0 \]

Using the quadratic formula again:

\[ y = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
\[ y = \frac{-8 \pm \sqrt{64 - 16}}{2} \]
\[ y = \frac{-8 \pm \sqrt{48}}{2} \]
\[ y = \frac{-8 \pm 4\sqrt{3}}{2} \]
\[ y = -4 \pm 2\sqrt{3} \]

### Step 8: Find \(x\) values

Now, we need to find the values of \(x\) for \(y = -4 + 2\sqrt{3}\) and \(y = -4 - 2\sqrt{3}\).

### Step 9: Determine the number of solutions

The tangent function has a period of \(\pi\), and each value of \(y\) will correspond to two solutions in the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\) and will repeat every \(\pi\). Therefore, for each valid \(y\), there are two solutions in the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\) and infinitely many solutions in the real line.

Thus, the total number of real solutions is infinite.

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