Question 37

The values of $x$ , for $0° \leq x \leq 360°$ that satisfy $\sin x + \sin 2x > \sin 3x$ are....

$0° < x < 120°, 180° < x < 240°$

$0° < x < 150°, 180° < x < 270°$

$120° < x < 180°, 240° < x < 360°$

$150° < x < 180°, 270° < x < 360°$

$0° < x < 135°, 180° < x < 270°$

Explanation

Given the form of the inequality

\sin x + \sin 2x > \sin 3x

Subtract $\sin x$ from both sides

\sin 2x > \sin 3x - \sin x

Use trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}$

2 \sin x \cos x > 2 \cos \frac{3x + x}{2} \sin \frac{3x - x}{2}

2 \sin x \cos x > 2 \cos 2x \sin x

Divide both sides by $2$ and use the identity $\cos 2x = 2 \cos^2 x - 1$

\sin x \cos x > \cos 2x \sin x

\sin x \cos x > (2 \cos^2 x - 1) \sin x

\sin x \cos x - (2 \cos^2 x - 1) \sin x > 0

\sin x (\cos x - 2 \cos^2 x + 1) > 0

\sin x (2 \cos^2 x - \cos x - 1) < 0

Factor $2 \cos^2 x - \cos x - 1 = (2 \cos x + 1)(\cos x - 1)$

\sin x (2 \cos x + 1)(\cos x - 1) < 0

Thus, the zeros are

\sin x = 0 \Rightarrow x = 0°, 180°, 360°

2 \cos x + 1 = 0 \Rightarrow \cos x = -\frac{1}{2} \Rightarrow x = 120°, 240°

\cos x - 1 = 0 \Rightarrow \cos x = 1 \Rightarrow x = 0°, 360°

The critical points are $x = 0°$ , $x = 120°$ , $x = 180°$ , $x = 240°$ , and $x = 360°$ .

The number line is as follows

Number Line

Critical points:

0°

120°

180°

240°

, and

360°

\text{Interval 1}

\text{Interval 2}

\text{Interval 3}

\text{Interval 4}

0°

120°

120°

180°

180°

240°

240°

360°

By testing the sign on each interval, we obtain that the inequality $\sin x (2 \cos x + 1)(\cos x - 1) < 0$ is satisfied for $0° < x < 120°$ and $180° < x < 240°$ .

So the solution set is $0° < x < 120°, 180° < x < 240°$ .

Command Palette

Number 37

Explanation