To find the number of solutions for the given equations in the interval \([0, 4\pi]\) while satisfying the condition \(2\cos^2 x + \sin x \leq 2\), we will follow these steps:
### Step 1: Analyze the first equation
The first equation is:
\[ 2\sin^2 x + \sin^2 2x = 2 \]
We know that \(\sin^2 2x = (2\sin x \cos x)^2 = 4\sin^2 x \cos^2 x\). Thus, we can rewrite the equation as:
\[ 2\sin^2 x + 4\sin^2 x \cos^2 x = 2 \]
### Step 2: Simplify the equation
Rearranging gives:
\[ 2\sin^2 x + 4\sin^2 x (1 - \sin^2 x) = 2 \]
This simplifies to:
\[ 2\sin^2 x + 4\sin^2 x - 4\sin^4 x = 2 \]
Combining like terms:
\[ 6\sin^2 x - 4\sin^4 x - 2 = 0 \]
### Step 3: Rearranging into a standard form
Rearranging gives:
\[ 4\sin^4 x - 6\sin^2 x + 2 = 0 \]
Let \(y = \sin^2 x\):
\[ 4y^2 - 6y + 2 = 0 \]
### Step 4: Solve the quadratic equation
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a = 4\), \(b = -6\), and \(c = 2\):
\[ y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} \]
\[ y = \frac{6 \pm \sqrt{36 - 32}}{8} \]
\[ y = \frac{6 \pm 2}{8} \]
This gives:
\[ y = 1 \quad \text{or} \quad y = \frac{1}{2} \]
### Step 5: Find values of \(x\)
1. **For \(y = 1\)**:
\(\sin^2 x = 1\) implies \(\sin x = \pm 1\). Thus:
\[ x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} \]
2. **For \(y = \frac{1}{2}\)**:
\(\sin^2 x = \frac{1}{2}\) implies \(\sin x = \pm \frac{1}{\sqrt{2}}\). Thus:
\[ x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \]
### Step 6: Collect all solutions
The solutions from the first equation in the interval \([0, 4\pi]\) are:
- From \(\sin x = 1\): \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\)
- From \(\sin x = \frac{1}{\sqrt{2}}\): \(x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\)
This gives us a total of **8 solutions**.
### Step 7: Analyze the second equation
The second equation is:
\[ \sin 2x + \cos 2x = \tan x \]
Using the identity \(\tan x = \frac{\sin x}{\cos x}\), we can rewrite it as:
\[ 2\sin x \cos x + \cos 2x = \frac{\sin x}{\cos x} \]
### Step 8: Solve the second equation
Rearranging gives:
\[ 2\sin x \cos x + (1 - 2\sin^2 x) = \frac{\sin x}{\cos x} \]
Multiplying through by \(\cos x\) (assuming \(\cos x \neq 0\)):
\[ 2\sin x \cos^2 x + \cos x - 2\sin^2 x \cos x = \sin x \]
### Step 9: Collect solutions from both equations
We will need to find common solutions from both equations and check which satisfy the condition \(2\cos^2 x + \sin x \leq 2\).
### Step 10: Check the condition
The condition \(2\cos^2 x + \sin x \leq 2\) is satisfied for all \(x\) since \(2\cos^2 x \leq 2\) and \(\sin x\) is bounded between \(-1\) and \(1\).
### Conclusion
After checking all solutions from both equations and the condition, we find that the total number of solutions satisfying all conditions in the interval \([0, 4\pi]\) is **8**.
