We are given the equation:

\[
7\sin^2A + 3\cos^2A = 4
\]

We need to find \(\tan A\).

### Step 1: Express \(\cos^2A\) in terms of \(\sin^2A\)
Since we know that:

\[
\sin^2A + \cos^2A = 1
\]

We can write:

\[
\cos^2A = 1 - \sin^2A
\]

### Step 2: Substitute \(\cos^2A\) into the original equation
Now, substitute \(\cos^2A = 1 - \sin^2A\) into the given equation:

\[
7\sin^2A + 3(1 - \sin^2A) = 4
\]

### Step 3: Simplify the equation
Expand and simplify:

\[
7\sin^2A + 3 - 3\sin^2A = 4
\]

Combine like terms:

\[
(7\sin^2A - 3\sin^2A) + 3 = 4
\]

\[
4\sin^2A + 3 = 4
\]

### Step 4: Solve for \(\sin^2A\)
Subtract 3 from both sides:

\[
4\sin^2A = 1
\]

Now, divide both sides by 4:

\[
\sin^2A = \frac{1}{4}
\]

### Step 5: Find \(\sin A\)
Taking the square root of both sides:

\[
\sin A = \frac{1}{2}
\]

### Step 6: Use \(\sin A\) to find \(\cos A\)
We know:

\[
\cos^2A = 1 - \sin^2A = 1 - \frac{1}{4} = \frac{3}{4}
\]

Taking the square root:

\[
\cos A = \frac{\sqrt{3}}{2}
\]

### Step 7: Find \(\tan A\)
Finally, use the definition of \(\tan A\):

\[
\tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
\]

Thus, \(\tan A = \frac{\sqrt{3}}{3}\).

---

Would you like more details on any of the steps? Here are some related questions to deepen understanding:

1. What happens if \(\cos A = -\frac{\sqrt{3}}{2}\)?
2. How can you use trigonometric identities to verify the result?
3. What is the general solution for \(A\) based on the values of \(\sin A\) and \(\cos A\)?
4. How does the Pythagorean identity help in solving trigonometric equations?
5. Can this problem be solved using a graphical method?

**Tip:** Whenever you encounter a trigonometric equation, try expressing all terms in one trigonometric function for simpler calculation!

Discover how to solve the trigonometric equation 7sin^2A + 3cos^2A = 4 to find tanA. This solution involves key trigonometric identities like sin^2A + cos^2A = 1 and helps students grasp the fundamentals of solving trigonometric equations. Suitable for high school students in Grades 10-12.

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