Solve the equation: 2x - 4 2x = 0 for 0^ x 180^.

$$1 = 4 \frac{\sin 2x}{\cos 2x}$$ $$1 = 4 \tan 2x$$ $$\tan 2x = \frac{1}{4}$$

Since $\tan 2x$ is positive, $2x$ can be in the first or third quadrant.

$$2x = 14.036^\circ + n \cdot 180^\circ \quad \text{where } n \text{ is an integer}$$

$$x = \frac{14.036^\circ}{2} + n \cdot \frac{180^\circ}{2}$$ $$x = 7.018^\circ + n \cdot 90^\circ$$

For $n=0$:

$$x = 7.018^\circ + 0 \cdot 90^\circ = 7.018^\circ$$

For $n=1$:

$$x = 7.018^\circ + 1 \cdot 90^\circ = 97.018^\circ$$

For $n=2$:

$$x = 7.018^\circ + 2 \cdot 90^\circ = 187.018^\circ \quad (\text{outside the range})$$

The final answers are $\boxed{7.02^\circ, 97.02^\circ}$ (rounded to two decimal places).

4. A team of 5 players is to be selected from 7 boys and 6 girls. If the team must have more girls than boys, in how many ways can the team be selected?
Let $B$ be the number of boys and $G$ be the number of girls.
Total players to be selected = 5.
Total boys available = 7.
Total girls available = 6.
Condition: $G > B$ and $B+G=5$.

We list the possible combinations of boys and girls that satisfy the conditions:

• Case 1: 2 boys and 3 girls ($2+3=5$, $3>2$) Number of ways = $\binom{7}{2} \times \binom{6}{3}$ $$\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21$$ $$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ Ways for Case 1 = $21 \times 20 = 420$
• Case 2: 1 boy and 4 girls ($1+4=5$, $4>1$) Number of ways = $\binom{7}{1} \times \binom{6}{4}$ $$\binom{7}{1} = 7$$ $$\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15$$ Ways for Case 2 = $7 \times 15 = 105$
• Case 3: 0 boys and 5 girls ($0+5=5$, $5>0$) Number of ways = $\binom{7}{0} \times \binom{6}{5}$ $$\binom{7}{0} = 1$$ $$\binom{6}{5} = \frac{6!}{5!(6-5)!} = \frac{6}{1} = 6$$ Ways for Case 3 = $1 \times 6 = 6$

Total number of ways = Sum of ways for all valid cases.
Total ways = $42