Find the total number of solutions of the equation sin4 x + cos4 x =

1.	Find the total number of solutions of the equation sin4 x + cos4 x = sin x cos x in [0, 2π].
Answer» Given sin⁴ x + cos⁴ x = sin x cos x ⇒ (sin² x + cos² x)² – 2 sin² x cos² x = sin x cos x ⇒ 1 - \(\frac{(2\,sinx\,cosx)^2}{2}\) = \(\frac{2\,sinx\,cosx}{2}\) ⇒ 1 – \(\frac{{sin}^2\,2x}{2}\) = \(\frac{sin\,2x}{2}\) ⇒ sin² 2x + sin 2x – 2 = 0 ⇒ (sin 2x + 2) (sin 2x – 1) = 0 ⇒ sin 2x = 1 ⇒ sin 2x = sin \(\frac{\pi}{2}\) = sin \(\big(\) 2π + \(\frac{\pi}{2}\) \(\big)\) ( ∵ sin 2x ≠ – 2 is indivisible) ⇒ 2x = \(\frac{\pi}{2}\) or \(\big(\) 2π + \(\frac{\pi}{2}\) \(\big)\) ⇒ 2x = \(\frac{\pi}{2}\) or \(\frac{5\pi}{2}\) ⇒ x = \(\frac{\pi}{4}\) or \(\frac{5\pi}{4}\)

Find the total number of solutions of the equation sin4 x + cos4 x = sin x cos x in [0, 2π].

Answer»

Given

sin⁴ x + cos⁴ x = sin x cos x

⇒ (sin² x + cos² x)² – 2 sin² x cos² x = sin x cos x

⇒ 1 - \(\frac{(2\,sinx\,cosx)^2}{2}\) = \(\frac{2\,sinx\,cosx}{2}\)

⇒ 1 – \(\frac{{sin}^2\,2x}{2}\) = \(\frac{sin\,2x}{2}\)

⇒ sin² 2x + sin 2x – 2 = 0

⇒ (sin 2x + 2) (sin 2x – 1) = 0

⇒ sin 2x = 1

⇒ sin 2x = sin \(\frac{\pi}{2}\) = sin \(\big(\) 2π + \(\frac{\pi}{2}\) \(\big)\) ( ∵ sin 2x ≠ – 2 is indivisible)

⇒ 2x = \(\frac{\pi}{2}\) or \(\big(\) 2π + \(\frac{\pi}{2}\) \(\big)\)

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

⇒ x = \(\frac{\pi}{4}\) or \(\frac{5\pi}{4}\)