$$ herefore x+30^{\circ}=60^{\circ}+k \cdot 360^{\circ} $$ or $$ x+30^{\circ}=120^{\circ}+k \cdot 360^{\circ} ; k \in \mathbb{Z} $$ $$ herefore x=30^{\circ}+k \cdot 360^{\circ} $$ or $$ x=90^{\circ}+k \cdot 360^{\circ} ; k \in \mathbb{Z} $$ EXERCISE 10 Solve for $$ x $$ : $$ \begin{array}{ll} ext { (a) } 3 \sin 2 x-2 \sin x=0 & ext { (b) } \cos 2 x+5 \cos x-2=0 ; x \in\left[0^{\circ} ; 360 ight.\end{array} $$

Question

$$ 	herefore x+30^{\circ}=60^{\circ}+k \cdot 360^{\circ} $$ or $$ x+30^{\circ}=120^{\circ}+k \cdot 360^{\circ} ; k \in \mathbb{Z} $$ $$ 	herefore x=30^{\circ}+k \cdot 360^{\circ} $$ or $$ x=90^{\circ}+k \cdot 360^{\circ} ; k \in \mathbb{Z} $$ EXERCISE 10 Solve for $$ x $$ : $$ \begin{array}{ll}	ext { (a) } 3 \sin 2 x-2 \sin x=0 & 	ext { (b) } \cos 2 x+5 \cos x-2=0 ; x \in\left[0^{\circ} ; 360ight.\end{array} $$

UpStudy AI · Accepted Answer

We are given two problems.

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Problem (a): Solve 3 sin 2x – 2 sin x = 0  
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Step 1. Write sin 2x in terms of sin x and cos x using the double‐angle formula:

sin 2x = 2 sin x cos x

Substitute into the equation:

3(2 sin x cos x) – 2 sin x = 0  
  => 6 sin x cos x – 2 sin x = 0

Step 2. Factor out 2 sin x:

2 sin x (3 cos x – 1) = 0

Step 3. Set each factor equal to zero:

• 2 sin x = 0     ⇒ sin x = 0  
• 3 cos x – 1 = 0   ⇒ cos x = 1/3

Step 4. Solve for x in the interval [0°, 360°).

For sin x = 0, we have:
  x = 0°  or  x = 180°

For cos x = 1/3, use the inverse cosine (note that 1/3 does not correspond to a common angle):
  x = arccos(1/3) ≈ 70.53°   
  and the cosine is positive in Quadrant I and IV so also:  
  x ≈ 360° – 70.53° ≈ 289.47°

Thus the solutions for (a) are:
  x = 0°, 70.53°, 180°, 289.47°.

─────────────────────────────  
Problem (b): Solve cos 2x + 5 cos x – 2 = 0  
─────────────────────────────

Step 1. Express cos 2x in terms of cos x using the double‐angle formula:
  cos 2x = 2 cos² x – 1

Substitute into the equation:

2 cos² x – 1 + 5 cos x – 2 = 0  
  => 2 cos² x + 5 cos x – 3 = 0

Step 2. Let u = cos x. Then the equation becomes:

2u² + 5u – 3 = 0

Step 3. Solve the quadratic equation using the quadratic formula:

u = [–5 ± √(5² – 4·2·(–3))] / (2·2)  
    = [–5 ± √(25 + 24)] / 4  
    = [–5 ± √49] / 4  
    = [–5 ± 7] / 4

This gives two roots:
  u = (–5 + 7)/4 = 2/4 = 1/2  
  u = (–5 – 7)/4 = –12/4 = –3

Since cos x must lie in [–1, 1] we reject u = –3.

Step 4. With cos x = 1/2, solve for x in [0°, 360°):

Cosine of 1/2 occurs at:
  x = 60°  and  x = 360° – 60° = 300°

Thus the solutions for (b) are:
  x = 60° and x = 300°.

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Final Answers:

(a) For 3 sin 2x – 2 sin x = 0, the solutions are:
  x = 0°, 70.53° (approximately), 180°, and 289.47° (approximately).

(b) For cos 2x + 5 cos x – 2 = 0, the solutions are:
  x = 60° and x = 300°.
Solutions for the equations are: (a) $$ x = 0^\circ, 70.53^\circ, 180^\circ, 289.47^\circ $$ (b) $$ x = 60^\circ, 300^\circ $$

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