Principal solutions of the equation `sin 2x + cos 2x = 0`, where `pi < x < 2pi `

Options

A. `(7pi)/(8),(11pi)/(8)`
B. `(9pi)/(8),(13pi)/(8)`
C. `(11pi)/(8),(15pi)/(8)`
D. `(15pi)/(8),(19pi)/(8)`

Answer

C

Explanation

Given equation in sin 2x+cos2x=0
`Rightarrow sin 2x=-cos 2x`
`Rightarrow tan 2x=-1" "[therefore pi t x lt 2pi Rightarrow 2pi lt 2x lt 4pi]`
`Rightarrow 2x=2pi +(3pi)/(4), 2pi+((3pi)/(2)+(pi)/(4))`
`Rightarrow 2x=(11x)/(8),(15pi)/(4)`
`Rightarrow x=(11pi)/(8), (15pi)/(8)`