Question

Find the exact solutions of the equation in the interval [0, 2π).

(4 sin x)(cos x)=1

Answer 1

By the double-angle formula for sine:
sin(2x) = 2sin(x)cos(x).

So:
4sin(x)cos(x) = 2sin(2x).

The equation now becomes:
4sin(x)cos(x) = 1
==> 2sin(2x) = 1
==> sin(2x) = 1/2.

This solves to get:
2x = π/6 ± 2πk and 2x = 5π/6 ± 2πk, where k is an integer
==> x = π/12 ± πk and x = 5π/12 ± πk.

All solutions that lie on [0, 2π) are given by k = 0 and k = 1. Therefore:
x = π/12, 5π/12, 13π/12, and 17π/12.

Answer 2

Where did the 13π/12 and 17π/12 come from?

Answer 3

the last equation that anonymus was looking at was

sin (2x) = 1/2
so from our basic triangles and the CAST rule
2x = π/6 , 2x = π-π/6 or 5π/6 , ( 30° or 150°)
then
x = π/12 or x = 5π/12

but the period of sin 2x is π
so by adding π to any answer already found we can get a new answer, so
π/12 + π = 13π/12 and 5π/12 = π = 17π/12

Answer 4

OH, alright. I understand now.
Thank you so much for your help!