Solve cos 3x + cos 2x = sin \(\frac{3}{2}\)x + sin\(\frac{1}{2}\)x ,�

1.	Solve cos 3x + cos 2x = sin \(\frac{3}{2}\)x + sin\(\frac{1}{2}\)x , 0 < x ≤ π .
Answer» cos 3x + cos 2x = sin\(\frac{3}{2}\)x + sin \(\frac{1}{2}\)x ⇒ 2 cos \(\frac{5}{2}\)x cos \(\frac{x}{2}\) = 2 sin x cos\(\frac{x}{2}\) ⇒ cos\(\frac{x}{2}\) \(\big[\) cos\(\frac{5x}{2}\) - sin x \(\big]\) = 0 ⇒ cos \(\frac{x}{2}\) = 0 or cos\(\frac{5x}{2}\) - sin x = 0 Now cos\(\frac{5x}{2}\) = 0 ⇒ \(\frac{x}{2}\) = \(\frac{\pi}{2}\) ⇒ x = π and cos\(\frac{5x}{2}\) - sin x = 0 ⇒ cos \(\frac{5x}{2}\) = sin x ⇒ cos \(\frac{5x}{2}\) = cos( \(\frac{\pi}{2}\) - x ) or sin (2π + \(\frac{\pi}{2}\) -x) ⇒ \(\frac{5x}{2}\) = \(\frac{\pi}{2}\) - x or, \(\frac{5x}{2}\) = 2π + \(\frac{\pi}{2}\) - x ⇒\(\frac{7}{2}\)x = \(\frac{π}{2}\) or \(\frac{7x}{2}\) = \(\frac{5π}{2}\) ⇒ x = \(\frac{\pi}{7}\) or \(\frac{5\pi}{7}\) ∴ x = \(\frac{\pi}{7}\) , \(\frac{5\pi}{7}\) or π

Solve cos 3x + cos 2x = sin \(\frac{3}{2}\)x + sin\(\frac{1}{2}\)x , 0 < x ≤ π .

Answer»

cos 3x + cos 2x = sin\(\frac{3}{2}\)x + sin \(\frac{1}{2}\)x

⇒ 2 cos \(\frac{5}{2}\)x cos \(\frac{x}{2}\) = 2 sin x cos\(\frac{x}{2}\)

⇒ cos\(\frac{x}{2}\) \(\big[\) cos\(\frac{5x}{2}\) - sin x \(\big]\) = 0

⇒ cos \(\frac{x}{2}\) = 0 or cos\(\frac{5x}{2}\) - sin x = 0

Now cos\(\frac{5x}{2}\) = 0 ⇒ \(\frac{x}{2}\) = \(\frac{\pi}{2}\) ⇒ x = π

and cos\(\frac{5x}{2}\) - sin x = 0

⇒ cos \(\frac{5x}{2}\) = sin x

⇒ cos \(\frac{5x}{2}\) = cos( \(\frac{\pi}{2}\) - x ) or sin (2π + \(\frac{\pi}{2}\) -x)

⇒ \(\frac{5x}{2}\) = \(\frac{\pi}{2}\) - x or, \(\frac{5x}{2}\) = 2π + \(\frac{\pi}{2}\) - x

⇒\(\frac{7}{2}\)x = \(\frac{π}{2}\) or \(\frac{7x}{2}\) = \(\frac{5π}{2}\)

⇒ x = \(\frac{\pi}{7}\) or \(\frac{5\pi}{7}\)

∴ x = \(\frac{\pi}{7}\) , \(\frac{5\pi}{7}\) or π