Solve : `sin^(10)x+cos^(10)x=29/16cos^4 2x`

Answer

To solve the equation \( \sin^{10} x + \cos^{10} x = \frac{29}{16} \cos^4 2x \), we can follow these steps:

### Step 1: Rewrite the Equation
We start by rewriting the left-hand side using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \sin^{10} x + \cos^{10} x = (\sin^2 x)^5 + (\cos^2 x)^5 \]
Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Then \( a + b = 1 \), and we can express the left-hand side as:
\[ a^5 + b^5 \]

### Step 2: Use the Identity for Sums of Powers
Using the identity \( a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4) \):
\[ a^5 + b^5 = 1 \cdot (a^4 - a^3b + a^2b^2 - ab^3 + b^4) = a^4 - a^3b + a^2b^2 - ab^3 + b^4 \]

### Step 3: Express \( \cos^4 2x \)
Next, we express \( \cos^4 2x \) in terms of \( \sin^2 x \) and \( \cos^2 x \):
\[ \cos 2x = \cos^2 x - \sin^2 x = b - a \]
Thus,
\[ \cos^4 2x = (b - a)^4 \]

### Step 4: Substitute and Rearrange
Now we substitute back into the original equation:
\[ a^5 + b^5 = \frac{29}{16} (b - a)^4 \]
This gives us:
\[ a^5 + b^5 = \frac{29}{16} ((b - a)^4) \]

### Step 5: Let \( t = \cos 2x \)
Let \( t = \cos 2x \). Then we can express \( a \) and \( b \) in terms of \( t \):
\[ a = \frac{1 - t}{2}, \quad b = \frac{1 + t}{2} \]
Substituting these into the equation gives us:
\[ \left(\frac{1 - t}{2}\right)^5 + \left(\frac{1 + t}{2}\right)^5 = \frac{29}{16} t^4 \]

### Step 6: Simplify the Equation
Now we simplify the left-hand side:
\[ \frac{(1 - t)^5 + (1 + t)^5}{32} = \frac{29}{16} t^4 \]
Multiplying through by 32 gives:
\[ (1 - t)^5 + (1 + t)^5 = 58 t^4 \]

### Step 7: Expand Using Binomial Theorem
Using the binomial theorem:
\[ (1 - t)^5 = 1 - 5t + 10t^2 - 10t^3 + 5t^4 - t^5 \]
\[ (1 + t)^5 = 1 + 5t + 10t^2 + 10t^3 + 5t^4 + t^5 \]
Adding these:
\[ 2 + 20t^2 + 10t^4 = 58t^4 \]
This simplifies to:
\[ 2 + 20t^2 - 48t^4 = 0 \]

### Step 8: Rearranging into Standard Form
Rearranging gives us:
\[ 48t^4 - 20t^2 - 2 = 0 \]

### Step 9: Let \( u = t^2 \)
Let \( u = t^2 \):
\[ 48u^2 - 20u - 2 = 0 \]

### Step 10: Solve the Quadratic Equation
Using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 48 \cdot (-2)}}{2 \cdot 48} \]
Calculating the discriminant:
\[ 400 + 384 = 784 \quad \Rightarrow \quad \sqrt{784} = 28 \]
Thus:
\[ u = \frac{20 \pm 28}{96} \]
Calculating the two possible values for \( u \):
1. \( u = \frac{48}{96} = \frac{1}{2} \)
2. \( u = \frac{-8}{96} \) (not valid since \( u \) cannot be negative)

### Step 11: Find \( t \)
Since \( u = t^2 \), we have:
\[ t^2 = \frac{1}{2} \quad \Rightarrow \quad t = \pm \frac{1}{\sqrt{2}} \]

### Step 12: Solve for \( x \)
Recall that \( t = \cos 2x \):
\[ \cos 2x = \frac{1}{\sqrt{2}} \quad \text{or} \quad \cos 2x = -\frac{1}{\sqrt{2}} \]
This gives:
1. \( 2x = \frac{\pi}{4} + 2n\pi \) or \( 2x = -\frac{\pi}{4} + 2n\pi \)
2. \( 2x = \frac{3\pi}{4} + 2n\pi \) or \( 2x = -\frac{3\pi}{4} + 2n\pi \)

Dividing by 2:
\[ x = \frac{\pi}{8} + n\pi \quad \text{or} \quad x = \frac{3\pi}{8} + n\pi \]

### Final Solution
Thus, the solutions for \( x \) are:
\[ x = n\pi + \frac{\pi}{8} \quad \text{and} \quad x = n\pi + \frac{3\pi}{8}, \quad n \in \mathbb{Z} \]