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 4 - 2011 - Paper 4

Next, rearranging gives us a quadratic in terms of sine:

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

or,

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

Applying the quadratic formula, where ( a = 4, b = -2, c = -1 ):

$\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(4)(-1)}}{2(4)}$

This results in:

$\sin \theta = \frac{2 \pm \sqrt{4 + 16}}{8} = \frac{2 \pm \sqrt{20}}{8} = \frac{2 \pm 2\sqrt{5}}{8} = \frac{1 \pm \sqrt{5}}{4}.$

The principal values can be derived from the solutions of:

• For ( \sin \theta = \frac{1 + \sqrt{5}}{4} ): Calculate the angles.
• For ( \sin \theta = \frac{1 - \sqrt{5}}{4} ): Since this yields a negative sine value, find angles for sine being negative.

Calculating:

• The solutions yield ( \theta = 54^\circ, 126^\circ, 198^\circ, 342^\circ ).