Find all the solutions, in the interval $0 \leq x < 2\pi$, of the equation $$2 \cos^2 x + 1 = 5 \sin x,$$ giving each solution in terms of $\pi$. - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 2

Next, we can factor the quadratic equation:

$-2\sin^2 x - 5 \sin x + 3 = 0$

Multiplying through by -1 to simplify:

$2\sin^2 x + 5 \sin x - 3 = 0.$

Now, we can factor it:

$(2\sin x - 1)(\sin x + 3) = 0.$

Since (\sin x + 3 = 0) has no real solutions (as sin is bounded between -1 and 1), we focus on the first factor.

Setting (2\sin x - 1 = 0), we find:

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

The solutions for this within the interval (0 \leq x < 2\pi) are:

$x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}.$

Thus, the solutions in terms of (\pi) are:

• (x = \frac{\pi}{6})
• (x = \frac{5\pi}{6})