Find the number of solutions of the equation `sin^4x+cos^4x-2sin^2x+3/4sin^2 2x=0` in the interval `[0,2pi]`

Explanation

To find the number of solutions of the equation

\[ \sin^4 x + \cos^4 x - 2\sin^2 x + \frac{3}{4} \sin^2 2x = 0 \]

in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Rewrite the equation using identities
We start by using the identity for \(\sin^2 2x\):
\[ \sin^2 2x = 4\sin^2 x \cos^2 x \]
Substituting this into the equation gives:
\[ \sin^4 x + \cos^4 x - 2\sin^2 x + \frac{3}{4} \cdot 4\sin^2 x \cos^2 x = 0 \]
This simplifies to:
\[ \sin^4 x + \cos^4 x - 2\sin^2 x + 3\sin^2 x \cos^2 x = 0 \]

### Step 2: Use the identity for \(\sin^4 x + \cos^4 x\)
We can use the identity:
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Substituting this into our equation gives:
\[ 1 - 2\sin^2 x \cos^2 x - 2\sin^2 x + 3\sin^2 x \cos^2 x = 0 \]
Combining like terms results in:
\[ 1 + \sin^2 x \cos^2 x - 2\sin^2 x = 0 \]

### Step 3: Substitute \(\sin^2 x\)
Let \(y = \sin^2 x\). Then \(\cos^2 x = 1 - y\), and we can rewrite the equation:
\[ 1 + y(1 - y) - 2y = 0 \]
This simplifies to:
\[ 1 + y - y^2 - 2y = 0 \implies -y^2 - y + 1 = 0 \]
Multiplying through by -1 gives:
\[ y^2 + y - 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]
This gives us two potential solutions for \(y\):
\[ y_1 = \frac{-1 + \sqrt{5}}{2}, \quad y_2 = \frac{-1 - \sqrt{5}}{2} \]

### Step 5: Determine valid solutions for \(y\)
Since \(y = \sin^2 x\), it must be in the range \([0, 1]\).

1. For \(y_1 = \frac{-1 + \sqrt{5}}{2}\):
- Calculate \(\sqrt{5} \approx 2.236\), so \(y_1 \approx \frac{-1 + 2.236}{2} \approx 0.618\) (valid).

2. For \(y_2 = \frac{-1 - \sqrt{5}}{2}\):
- This value is negative, hence not valid.

### Step 6: Find solutions for valid \(y\)
Now we have \(y = \sin^2 x = \frac{-1 + \sqrt{5}}{2}\).

To find \(x\):
\[ \sin x = \pm \sqrt{\frac{-1 + \sqrt{5}}{2}} \]
This gives us two solutions in the interval \([0, 2\pi]\) for \(\sin x = \sqrt{\frac{-1 + \sqrt{5}}{2}}\) and two solutions for \(\sin x = -\sqrt{\frac{-1 + \sqrt{5}}{2}}\).

### Conclusion
Thus, there are a total of **4 solutions** in the interval \([0, 2\pi]\).

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