If $$ 5 \cos x+3=0 $$ and $$ 180

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To find the value of $$ 3 \tan x + 25 \sin^2 x $$ given that $$ 5 \cos x + 3 = 0 $$ and $$ 180^\circ < x < 300^\circ $$, follow these steps:

### Step 1: Solve for $$ \cos x $$

Given:
$$
5 \cos x + 3 = 0
$$
$$
\Rightarrow 5 \cos x = -3
$$
$$
\Rightarrow \cos x = -\frac{3}{5}
$$

### Step 2: Determine the Quadrant

Since $$ 180^\circ < x < 300^\circ $$:

- $$ 180^\circ < x < 270^\circ $$ lies in **Quadrant III**.
- $$ 270^\circ < x < 300^\circ $$ lies in **Quadrant IV**.

Given $$ \cos x = -\frac{3}{5} $$ (negative), $$ x $$ must be in **Quadrant III**, where both sine and cosine are negative.

### Step 3: Find $$ \sin x $$

Use the Pythagorean identity:
$$
\sin^2 x + \cos^2 x = 1
$$
$$
\sin^2 x = 1 - \cos^2 x = 1 - \left( -\frac{3}{5} \right)^2 = 1 - \frac{9}{25} = \frac{16}{25}
$$
Since $$ x $$ is in Quadrant III, $$ \sin x $$ is negative:
$$
\sin x = -\frac{4}{5}
$$

### Step 4: Calculate $$ \tan x $$

$$
\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}
$$

### Step 5: Compute $$ 3 \tan x + 25 \sin^2 x $$

$$
3 \tan x + 25 \sin^2 x = 3 \left( \frac{4}{3} \right) + 25 \left( \frac{16}{25} \right) = 4 + 16 = 20
$$

### **Final Answer:**
The value of $$ 3 \tan x + 25 \sin^2 x $$ is **20**.
The value of $$ 3 \tan x + 25 \sin^2 x $$ is **20**.

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