Find all the solutions of $$2 \cos 2\theta = 1 - 2 \sin \theta$$ in the interval $0 \leq \theta < 360^\circ$. - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 4

This simplifies to:

$2 - 4\sin^2 \theta = 1 - 2\sin \theta$

Rearranging gives us:

$4\sin^2 \theta - 2\sin \theta - 1 = 0$

We can treat this as a standard quadratic equation in the form:

$a\sin^2 \theta + b\sin \theta + c = 0$

Identifying the coefficients: $a = 4, b = -2, c = -1$

We apply the quadratic formula:

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

Calculating the discriminant:

$D = (-2)^2 - 4 \cdot 4 \cdot (-1) = 4 + 16 = 20$

Now substituting into the formula:

$\sin \theta = \frac{2 \pm \sqrt{20}}{8}$

We simplify:

$\sin \theta = \frac{2 \pm 2\sqrt{5}}{8} = \frac{1 \pm \sqrt{5}}{4}$

This gives us two possible values for \sin \theta:

1. (\sin \theta = \frac{1 + \sqrt{5}}{4})
2. (\sin \theta = \frac{1 - \sqrt{5}}{4})

We will only consider values within the range $0 \leq \theta < 360^\circ$.

For each sine value, we determine angles:

1. For (\sin \theta = \frac{1 + \sqrt{5}}{4}):

   • This will give angles in quadrants 1 and 2.

2. For (\sin \theta = \frac{1 - \sqrt{5}}{4}):

   • Check if this value is valid (remains between -1 and 1).
   • The angles can also be derived from the reference angle accordingly considering sine's properties.

Using inverse sine for valid values, the solutions yield:

$\theta = \{54^\circ, 126^\circ, 198^\circ, 342^\circ\}$