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WAEC Further Mathematics 2013 Paper

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Question : 30

Total: 40

If

2\sin^{2}\theta = 1 + \cos\theta,\quad 0^{\circ} \le \theta \le 90^{\circ}

\theta

.

30°
45°
60°
90°

Solution:

Concept:

Use

\sin^2\theta = 1 - \cos^2\theta

to form a quadratic equation in

\cos\theta

.

Explanation:

Given:

2\sin^2\theta = 1 + \cos\theta

.

Replace

\sin^2\theta

with

1 - \cos^2\theta

.

So,

2(1 - \cos^2\theta) = 1 + \cos\theta

.

Simplify:

2 - 2\cos^2\theta = 1 + \cos\theta

.

Rearrange:

2\cos^2\theta + \cos\theta - 1 = 0

.

Factorize:

(2\cos\theta - 1)(\cos\theta + 1) = 0

.

Thus,

\cos\theta = \frac{1}{2}

or

\cos\theta = -1

.

Since

0^\circ \leq \theta \leq 90^\circ

, reject

\cos\theta = -1

.

Hence,

\cos\theta = \frac{1}{2}

, giving

\theta = 60^\circ

.

Answer:

\theta = 60^\circ

, which is option C.

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