Consider the equation `5 sin^2 x + 3 sin x cos x - 3 cos^2 x =2 `.......... (i)
`sin^2 x - cos 2 x =2-sin 2 x `........... (ii)
The number of solutions common to (i) and (ii) is 0 1 finite infinite

Options

A. 0
B. 1
C. finite
D. infinite

Answer

To solve the problem, we need to analyze the two equations provided and determine the number of common solutions.

### Step 1: Analyze the first equation
The first equation is given as:
\[ 5 \sin^2 x + 3 \sin x \cos x - 3 \cos^2 x = 2 \]

We can rewrite this equation in terms of \( \tan x \). Using the identities \( \sin^2 x = \tan^2 x \cos^2 x \) and \( \cos^2 x = \frac{1}{1 + \tan^2 x} \), we can express everything in terms of \( \tan x \).

### Step 2: Substitute and rearrange
Dividing the entire equation by \( \cos^2 x \) gives:
\[ 5 \tan^2 x + 3 \tan x - 3 = 2 \sec^2 x \]

Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can rewrite the equation:
\[ 5 \tan^2 x + 3 \tan x - 3 = 2(1 + \tan^2 x) \]
This simplifies to:
\[ 5 \tan^2 x + 3 \tan x - 3 = 2 + 2 \tan^2 x \]
Rearranging gives:
\[ 3 \tan^2 x + 3 \tan x - 5 = 0 \]

### Step 3: Solve the quadratic equation
Let \( t = \tan x \). The equation becomes:
\[ 3t^2 + 3t - 5 = 0 \]

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = 3, c = -5 \):
\[ t = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \]
\[ t = \frac{-3 \pm \sqrt{9 + 60}}{6} \]
\[ t = \frac{-3 \pm \sqrt{69}}{6} \]

### Step 4: Analyze the second equation
The second equation is:
\[ \sin^2 x - \cos 2x = 2 - \sin 2x \]

Using the identity \( \cos 2x = 1 - 2\sin^2 x \) and \( \sin 2x = 2 \sin x \cos x \), we can rewrite the equation:
\[ \sin^2 x - (1 - 2\sin^2 x) = 2 - 2 \sin x \cos x \]
This simplifies to:
\[ 3\sin^2 x - 1 = 2 - 2\sin x \cos x \]

### Step 5: Rearranging the second equation
Rearranging gives:
\[ 3\sin^2 x + 2\sin x \cos x - 3 = 0 \]

### Step 6: Solve the second equation
Dividing by \( \cos^2 x \):
\[ 3\tan^2 x + 2\tan x - 3 = 0 \]

Using the quadratic formula again:
Let \( u = \tan x \):
\[ 3u^2 + 2u - 3 = 0 \]
Using the quadratic formula:
\[ u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \]
\[ u = \frac{-2 \pm \sqrt{4 + 36}}{6} \]
\[ u = \frac{-2 \pm \sqrt{40}}{6} \]
\[ u = \frac{-2 \pm 2\sqrt{10}}{6} \]
\[ u = \frac{-1 \pm \sqrt{10}}{3} \]

### Step 7: Compare the solutions
Now we have two sets of solutions for \( \tan x \):
1. From the first equation: \( t = \frac{-3 \pm \sqrt{69}}{6} \)
2. From the second equation: \( u = \frac{-1 \pm \sqrt{10}}{3} \)

Since the values of \( t \) and \( u \) are different, there are no common solutions.

### Conclusion
The number of solutions common to both equations is **0**.

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