The general solution of ` 4sin^(4) x + cos^(4) x=1 ` is

Options

A. `(2n+1)""(pi)/(2)`
B. `npi`
C. `npi pm sin^(-1)sqrt((2)/(5))`
D. None of these

Answer

To find the general solution of the equation \( 4\sin^4 x + \cos^4 x = 1 \), we can 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 \(\sin^4 x\):
\[ 4\sin^4 x = 1 - \cos^4 x \]

### Step 2: Use the Pythagorean identity
Recall that \(\cos^2 x = 1 - \sin^2 x\). Therefore, we can rewrite \(\cos^4 x\) as:
\[ \cos^4 x = (1 - \sin^2 x)^2 \]
Substituting this into the equation gives:
\[ 4\sin^4 x + (1 - \sin^2 x)^2 = 1 \]

### Step 3: Expand the equation
Now, expand \((1 - \sin^2 x)^2\):
\[ (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x \]
Substituting this back into our equation yields:
\[ 4\sin^4 x + 1 - 2\sin^2 x + \sin^4 x = 1 \]
Combining like terms results in:
\[ 5\sin^4 x - 2\sin^2 x = 0 \]

### Step 4: Factor the equation
We can factor out \(\sin^2 x\):
\[ \sin^2 x (5\sin^2 x - 2) = 0 \]

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

For the first case:
\[ \sin^2 x = 0 \implies \sin x = 0 \implies x = n\pi, \quad n \in \mathbb{Z} \]

For the second case:
\[ 5\sin^2 x = 2 \implies \sin^2 x = \frac{2}{5} \implies \sin x = \pm \sqrt{\frac{2}{5}} \]

### Step 6: Solve for \(x\) in the second case
This gives us:
\[ x = n\pi + \sin^{-1}\left(\sqrt{\frac{2}{5}}\right) \quad \text{and} \quad x = n\pi - \sin^{-1}\left(\sqrt{\frac{2}{5}}\right) \]

### Final General Solution
Combining both cases, the general solution is:
\[ x = n\pi \quad \text{and} \quad x = n\pi \pm \sin^{-1}\left(\sqrt{\frac{2}{5}}\right), \quad n \in \mathbb{Z} \]