The solution set of sin \(\big(\) x + \(\frac{π}{4}\) \(\big)\)

1.	The solution set of sin \(\big(\) x + \(\frac{π}{4}\) \(\big)\) = sin 2x equals (a) \(\frac{n\pi+ \pi/4}{1-(-1)^n \,2}\)(b) \(\frac{n\pi- \pi/4}{1-(-1)^n \,2}\) (c) \(\frac{n\pi+ \pi/4}{1+ (-1)^n \,2}\) (d) \(\frac{n\pi- \pi/4}{1+ (-1)^n \,2}\)
Answer» Answer : (b) \(\frac{n\pi- \pi/4}{1-(-1)^n \,2}\) sin (x + π/4) = sin 2x ⇒ x + π/4 = nπ + (–1)n 2x, n ∈ I ⇒ x – (–1)ⁿ 2x = nπ – π/4, n ∈ I ⇒ x {1 – (–1)ⁿ .2} = nπ – π/4 ⇒ x = \(\frac{n\pi- \pi/4}{1-(-1)^n \,2}\) , n ∈ I.

The solution set of sin \(\big(\) x + \(\frac{π}{4}\) \(\big)\) = sin 2x equals (a) \(\frac{n\pi+ \pi/4}{1-(-1)^n \,2}\)(b) \(\frac{n\pi- \pi/4}{1-(-1)^n \,2}\) (c) \(\frac{n\pi+ \pi/4}{1+ (-1)^n \,2}\) (d) \(\frac{n\pi- \pi/4}{1+ (-1)^n \,2}\)

Answer»

Answer : (b) \(\frac{n\pi- \pi/4}{1-(-1)^n \,2}\)

sin (x + π/4) = sin 2x

⇒ x + π/4 = nπ + (–1)n 2x, n ∈ I

⇒ x – (–1)ⁿ 2x = nπ – π/4, n ∈ I

⇒ x {1 – (–1)ⁿ .2} = nπ – π/4

⇒ x = \(\frac{n\pi- \pi/4}{1-(-1)^n \,2}\) , n ∈ I.