For the interval 0^\circ \le x \le 360^\circ, solve the trigonometric equation 2\cos^2 x -… — JAMB UTME

For the interval $0^\circ \le x \le 360^\circ$ , solve the trigonometric equation $2\cos^2 x - \cos x - 1 = 0$ . Which set contains all the solutions for $x$ ?

$0^\circ, 120^\circ, 240^\circ, 360^\circ$ Answer
B
$60^\circ, 120^\circ, 240^\circ, 300^\circ$
C
$0^\circ, 120^\circ, 360^\circ$
D
$60^\circ, 300^\circ, 360^\circ$

Answer

The complete set of solutions is

0^\circ, 120^\circ, 240^\circ, 360^\circ

Factoring

2\cos^2 x - \cos x - 1 = 0

gives

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

. Setting the first factor to zero yields

\cos x = -\frac{1}{2}

, which has solutions at

120^\circ

and

240^\circ

in the interval

[0^\circ, 360^\circ]

. Setting the second factor to zero gives

\cos x = 1

, which has solutions at

0^\circ

and

360^\circ

. Combining these yields the set

0^\circ, 120^\circ, 240^\circ, 360^\circ

Step-by-Step Solution

Factor the quadratic trigonometric equation

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

Treat

\cos x

as a single variable to simplify into standard quadratic factors.

Set each factor to zero to find values for

\cos x

\cos x = 1

\cos x = -\frac{1}{2}

Zero-product property requires at least one factor to be zero.

Solve for

x

in the interval

0^\circ \le x \le 360^\circ

For

\cos x = 1

x = 0^\circ, 360^\circ

. For

\cos x = -\frac{1}{2}

x = 180^\circ - 60^\circ = 120^\circ

(Quadrant II) and

x = 180^\circ + 60^\circ = 240^\circ

(Quadrant III).

Cosine is negative in Quadrants II and III, with a reference angle of

60^\circ

Combine all unique solutions in ascending order

x = 0^\circ, 120^\circ, 240^\circ, 360^\circ

Include all solutions within the given domain boundaries.

Key Concept

Solving Quadratic Trigonometric Equations

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