How do you find the exact solutions of the equation #4sinxcosx=1# in the interval #[0,2pi)#?

Question

10437 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-27T10:11:34

#x = 1/12 pi, 5/12 pi, 13/12 pi and 17/12 pi#

#4sin x cos x = 1#

#2 ( 2 sin x cos x) =1#, where #2sin x cos x = sin 2 x#

# 2 sin 2x = 1#

#sin 2x = 1/2#

#2x =sin^(-1)(1/2)#

#2x = 1/6 pi,5/6 pi, 13/6 pi, 17/6 pi#

#x = 1/12 pi, 5/12 pi, 13/12 pi and 17/12 pi#