The largest positive solution of `1+"sin"^(4) x = "cos"^(2) 3x " in " [-5pi//2, 5pi//2]`, is

Options

A. `pi`
B. `2pi`
C. `(5pi)/(2)`
D. none of these

Answer

To solve the equation \(1 + \sin^4 x = \cos^2 3x\) in the interval \([-5\pi/2, 5\pi/2]\) and find the largest positive solution, we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ 1 + \sin^4 x = \cos^2 3x \]
Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we can rewrite \(\cos^2 3x\):
\[ 1 + \sin^4 x = 1 - \sin^2 3x \]
This simplifies to:
\[ \sin^4 x + \sin^2 3x = 0 \]

### Step 2: Express \(\sin^2 3x\)
Recall the triple angle formula for sine:
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
Thus,
\[ \sin^2 3x = (3 \sin x - 4 \sin^3 x)^2 \]

### Step 3: Substitute and simplify
Substituting \(\sin^2 3x\) into our equation gives:
\[ \sin^4 x + (3 \sin x - 4 \sin^3 x)^2 = 0 \]
Expanding the square:
\[ \sin^4 x + (9 \sin^2 x - 24 \sin^4 x + 16 \sin^6 x) = 0 \]
Combining like terms:
\[ 16 \sin^6 x - 23 \sin^4 x + 9 \sin^2 x = 0 \]

### Step 4: Factor out \(\sin^2 x\)
Factoring out \(\sin^2 x\):
\[ \sin^2 x (16 \sin^4 x - 23 \sin^2 x + 9) = 0 \]
This gives us two cases:
1. \(\sin^2 x = 0\)
2. \(16 \sin^4 x - 23 \sin^2 x + 9 = 0\)

### Step 5: Solve \(\sin^2 x = 0\)
The solutions for \(\sin^2 x = 0\) are:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]
In the interval \([-5\pi/2, 5\pi/2]\), the solutions are:
\[ x = -5\pi, -4\pi, -3\pi, -2\pi, -\pi, 0, \pi, 2\pi, 3\pi, 4\pi, 5\pi \]
The largest positive solution here is \(5\pi\).

### Step 6: Solve the quadratic equation
Now, solve the quadratic equation:
\[ 16y^2 - 23y + 9 = 0 \quad \text{where } y = \sin^2 x \]
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 16 \cdot 9}}{2 \cdot 16} \]
Calculating the discriminant:
\[ 23^2 - 4 \cdot 16 \cdot 9 = 529 - 576 = -47 \]
Since the discriminant is negative, there are no real solutions from this quadratic.

### Conclusion
The only solutions come from \(\sin^2 x = 0\). Therefore, the largest positive solution in the interval \([-5\pi/2, 5\pi/2]\) is:
\[ \boxed{5\pi} \]