If `4sin^4x+cos^4x=1,t h e nxi se q u a lto(n in Z)` `npi` (b) `npi+-sin^(-1)sqrt(2/5)` `(2npi)/3` (d) `2npi+-pi/4`

Options

A. `n pi`
B. `npi pm sin^(-1) sqrt(2/5)`
C. `(2 n pi)/3`
D. `2n pi pm pi/4`

Explanation

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 given equation:
\[ 4\sin^4 x + \cos^4 x = 1 \]
We can express \(\cos^4 x\) in terms of \(\sin^4 x\):
\[ \cos^4 x = (1 - \sin^2 x)^2 \]
Thus, substituting this into the equation gives:
\[ 4\sin^4 x + (1 - \sin^2 x)^2 = 1 \]

### Step 2: Expand the equation
Now, 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:
\[ 4\sin^4 x + 1 - 2\sin^2 x + \sin^4 x = 1 \]
This simplifies to:
\[ 5\sin^4 x - 2\sin^2 x + 1 = 1 \]

### Step 3: Simplify the equation
Subtract 1 from both sides:
\[ 5\sin^4 x - 2\sin^2 x = 0 \]
Factoring out \(\sin^2 x\):
\[ \sin^2 x (5\sin^2 x - 2) = 0 \]

### Step 4: Solve for \(\sin^2 x\)
Setting each factor to zero gives us:
1. \(\sin^2 x = 0\)
2. \(5\sin^2 x - 2 = 0\)

From \(\sin^2 x = 0\):
\[ \sin x = 0 \implies x = n\pi, \quad n \in \mathbb{Z} \]

From \(5\sin^2 x - 2 = 0\):
\[ \sin^2 x = \frac{2}{5} \implies \sin x = \pm \sqrt{\frac{2}{5}} \]
Thus, the solutions for \(x\) are:
\[ x = \sin^{-1}\left(\sqrt{\frac{2}{5}}\right) + 2n\pi \quad \text{and} \quad x = \pi - \sin^{-1}\left(\sqrt{\frac{2}{5}}\right) + 2n\pi \]

### Step 5: Combine the solutions
The general solutions can be expressed as:
\[ x = n\pi \quad \text{or} \quad x = n\pi \pm \sin^{-1}\left(\sqrt{\frac{2}{5}}\right) \]

### Conclusion
Thus, the values of \(x\) that satisfy the equation are:
- \(x = n\pi\) (Option A)
- \(x = n\pi \pm \sin^{-1}\left(\sqrt{\frac{2}{5}}\right)\) (Option B)