Which set contains all the solutions to the trigonometric equation \sin 2x = \cos x for 0^… — JAMB UTME

Which set contains all the solutions to the trigonometric equation $\sin 2x = \cos x$ for $0^\circ \le x \le 180^\circ$ ?

The correct set of solutions is \{30^\circ, 90^\circ, 150^\circ\}.

Using the identity

\sin 2x = 2\sin x \cos x

, the equation becomes

2\sin x \cos x - \cos x = 0

. Factoring out

\cos x

gives

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

. Setting each factor to zero yields

\cos x = 0

(giving

x = 90^\circ

) and

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

(giving

x = 30^\circ

and

x = 150^\circ

). Thus, the complete set of solutions in the given interval is \{30^\circ, 90^\circ, 150^\circ\}.

Apply the double-angle identity for sine.

Substitute

\sin 2x = 2\sin x \cos x

into the equation to get

2\sin x \cos x = \cos x

This expresses the equation in terms of single angle

x

Rearrange and factor the equation.

2\sin x \cos x - \cos x = 0 \implies \cos x(2\sin x - 1) = 0

Factoring prevents losing valid roots that occur when a variable factor equals zero.

Set each factor to zero and solve for

x

in the interval

0^\circ \le x \le 180^\circ

First factor:

\cos x = 0 \implies x = 90^\circ

.
Second factor:

2\sin x - 1 = 0 \implies \sin x = \frac{1}{2} \implies x = 30^\circ

x = 180^\circ - 30^\circ = 150^\circ

Finding all principal and secondary angles within the specified domain.

Combine all valid solutions into a set.

x \in \{30^\circ, 90^\circ, 150^\circ\}

All three values satisfy the original equation and lie within

0^\circ \le x \le 180^\circ

Solving trigonometric equations using identities and factoring

Estimated Time:1m 30s

Question