One of the solutions of the equation 4 sin4 x + cos4 x = 1 is (a) nπ

1.	One of the solutions of the equation 4 sin4 x + cos4 x = 1 is (a) nπ (b) \(\frac{2nπ}{3}\)(c) \((n-1) \frac{\pi}{4}\) (d) \((2n+1)\frac{\pi}{2}\)
Answer» Answer : (a) nπ 4 sin⁴ x + cos⁴ x = 1 ⇒ 4 sin⁴ x + (1 – sin²x)²= 1 ⇒ 4 sin⁴x + 1 + sin⁴x – 2 sin²x = 1 ⇒ 5 sin⁴x – 2 sin²x = 0 ⇒ sin²x (5 sin²x – 2) = 0 ⇒ sin²x = 0 or 5 sin²x – 2 = 0 ⇒ sin x = 0 or sin x = \(\sqrt{\frac{2}{5}}\) ⇒ x = nπ or x = sin^–1 ( \(\sqrt{\frac{2}{5}}\) ) = nπ + (–1)ⁿsin^–1 ( \(\sqrt{\frac{2}{5}}\) ), n ∈ I ∴ x = nπ is one of the solutions of the given equation.

One of the solutions of the equation 4 sin4 x + cos4 x = 1 is (a) nπ (b) \(\frac{2nπ}{3}\)(c) \((n-1) \frac{\pi}{4}\) (d) \((2n+1)\frac{\pi}{2}\)

Answer»

Answer : (a) nπ

4 sin⁴ x + cos⁴ x = 1

⇒ 4 sin⁴ x + (1 – sin²x)²= 1

⇒ 4 sin⁴x + 1 + sin⁴x – 2 sin²x = 1

⇒ 5 sin⁴x – 2 sin²x = 0

⇒ sin²x (5 sin²x – 2) = 0

⇒ sin²x = 0 or 5 sin²x – 2 = 0

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

⇒ x = nπ or x = sin^–1 ( \(\sqrt{\frac{2}{5}}\) )

= nπ + (–1)ⁿsin^–1 ( \(\sqrt{\frac{2}{5}}\) ), n ∈ I

∴ x = nπ is one of the solutions of the given equation.