One of the general solutions of `4sin^4x+cos^4x=1` is

Options

A. `npi+-alpha/2,alpha=cos^(-1)(1/5),AAn in Z`
B. `npi+-alpha/2,alpha=cos^(-1)(3/5),AAn in Z`
C. `npi+-alpha/2,alpha=cos^(-1)(1/3),AAn in Z`
D. none of these

Answer

To solve the equation \( 4\sin^4 x + \cos^4 x = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 4\sin^4 x + \cos^4 x = 1 \]
We can express \(\cos^4 x\) in terms of \(\sin^2 x\):
\[ \cos^4 x = (1 - \sin^2 x)^2 \]
Substituting this into the equation gives:
\[ 4\sin^4 x + (1 - \sin^2 x)^2 = 1 \]

### Step 2: Expand the equation
Next, we expand \((1 - \sin^2 x)^2\):
\[ (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x \]
Substituting this back into the equation:
\[ 4\sin^4 x + 1 - 2\sin^2 x + \sin^4 x = 1 \]

### Step 3: Simplify the equation
Now, we simplify the equation:
\[ 4\sin^4 x + \sin^4 x - 2\sin^2 x + 1 - 1 = 0 \]
This simplifies to:
\[ 5\sin^4 x - 2\sin^2 x = 0 \]

### Step 4: Factor the equation
We can factor out \(\sin^2 x\):
\[ \sin^2 x(5\sin^2 x - 2) = 0 \]
This gives us two cases to solve:
1. \(\sin^2 x = 0\)
2. \(5\sin^2 x - 2 = 0\)

### Step 5: Solve the first case
For the first case:
\[ \sin^2 x = 0 \implies \sin x = 0 \]
The general solution for this is:
\[ x = n\pi, \quad n \in \mathbb{Z} \]

### Step 6: Solve the second case
For the second case:
\[ 5\sin^2 x = 2 \implies \sin^2 x = \frac{2}{5} \]
Taking the square root gives:
\[ \sin x = \pm \sqrt{\frac{2}{5}} \]
The general solutions for this are:
\[ x = \arcsin\left(\sqrt{\frac{2}{5}}\right) + 2n\pi \quad \text{and} \quad x = \pi - \arcsin\left(\sqrt{\frac{2}{5}}\right) + 2n\pi, \quad n \in \mathbb{Z} \]
and
\[ x = \arcsin\left(-\sqrt{\frac{2}{5}}\right) + 2n\pi \quad \text{and} \quad x = \pi - \arcsin\left(-\sqrt{\frac{2}{5}}\right) + 2n\pi, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, one of the general solutions of the equation \( 4\sin^4 x + \cos^4 x = 1 \) is:
\[ x = n\pi \quad \text{or} \quad x = \arcsin\left(\sqrt{\frac{2}{5}}\right) + 2n\pi, \quad n \in \mathbb{Z} \]