Find the general solution of `1+sin^3x + cos^(3)x=(3)/(2)sin 2x `

Answer

To find the general solution of the equation \(1 + \sin^3 x + \cos^3 x = \frac{3}{2} \sin 2x\), we can follow these steps:

### Step 1: Rewrite \( \sin^3 x + \cos^3 x \)
We can use the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = \sin x \) and \( b = \cos x \). Thus,
\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x) \]

### Step 2: Substitute into the original equation
Now, substituting this back into the original equation gives:
\[ 1 + (\sin x + \cos x)(1 - \sin x \cos x) = \frac{3}{2} \sin 2x \]
We know that \( \sin 2x = 2 \sin x \cos x \), so we can rewrite the equation as:
\[ 1 + (\sin x + \cos x)(1 - \sin x \cos x) = 3 \sin x \cos x \]

### Step 3: Simplify the equation
Rearranging gives:
\[ 1 + \sin x + \cos x - \sin x \cos x (\sin x + \cos x) = 3 \sin x \cos x \]
This simplifies to:
\[ 1 + \sin x + \cos x = 3 \sin x \cos x + \sin x \cos x (\sin x + \cos x) \]

### Step 4: Let \( y = \sin x + \cos x \)
Let \( y = \sin x + \cos x \). Then, we can express \( \sin x \cos x \) in terms of \( y \):
\[ \sin^2 x + \cos^2 x = 1 \Rightarrow \sin^2 x + \cos^2 x + 2 \sin x \cos x = y^2 \Rightarrow 1 + 2 \sin x \cos x = y^2 \]
Thus, we have:
\[ \sin x \cos x = \frac{y^2 - 1}{2} \]

### Step 5: Substitute back into the equation
Substituting \( \sin x \cos x \) back into our equation gives:
\[ 1 + y = 3 \cdot \frac{y^2 - 1}{2} + \frac{y^2 - 1}{2} y \]
This simplifies to:
\[ 1 + y = \frac{3y^2 - 3 + y^3 - y}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 2 + 2y = 3y^2 - 3 + y^3 - y \]
Rearranging gives:
\[ y^3 + 3y^2 - 3y - 5 = 0 \]

### Step 6: Factor the cubic equation
To find the roots of the cubic equation \( y^3 + 3y^2 - 3y - 5 = 0 \), we can test for rational roots. Testing \( y = 1 \):
\[ 1^3 + 3(1^2) - 3(1) - 5 = 1 + 3 - 3 - 5 = -4 \quad \text{(not a root)} \]
Testing \( y = -1 \):
\[ (-1)^3 + 3(-1)^2 - 3(-1) - 5 = -1 + 3 + 3 - 5 = 0 \quad \text{(is a root)} \]
So, \( y + 1 \) is a factor. Dividing \( y^3 + 3y^2 - 3y - 5 \) by \( y + 1 \) gives:
\[ y^2 + 2y - 5 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6} \]

### Step 8: Find \( \sin x + \cos x \)
Thus, we have:
\[ \sin x + \cos x = -1, \quad \sin x + \cos x = -1 + \sqrt{6}, \quad \sin x + \cos x = -1 - \sqrt{6} \]

### Step 9: Solve for \( x \)
Using \( \sin x + \cos x = k \), we can express it as:
\[ \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = k \]
Thus, we can find \( x \) for each case.