What is the general solution of the equation $2\sin^2 x - 7\sin x + 3 = 0$? (A) $n \pi - (-1)^n \frac{\pi}{3}$ (B) $n \pi + (-1)^n \frac{\pi}{3}$ (C) $n \pi - (-1)^n \frac{\pi}{6}$ (D) $n \pi + (-1)^n \frac{\pi}{6}$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2016 - Paper 1

To solve the quadratic equation, we use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case:

• (a = 2)
• (b = -7)
• (c = 3)

Calculating the discriminant:

$b^2 - 4ac = (-7)^2 - 4 \times 2 \times 3 = 49 - 24 = 25$

Now substituting values into the formula:

$y = \frac{7 \pm 5}{4}$

This gives us two potential solutions:

1. (y = \frac{12}{4} = 3)
2. (y = \frac{2}{4} = \frac{1}{2})

Since (y = \sin x), we disregard (y = 3) as it is not a valid sine value. Therefore, we have:

$\sin x = \frac{1}{2}$

The angle that satisfies (\sin x = \frac{1}{2}) is:

$x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi$

where (k) is any integer. Thus, combining these solutions leads us to the general solution:

$x = n\pi + (-1)^n \frac{\pi}{6}$

This corresponds to option (D) in the original question.