Question

How do you solve `sin2x + 2 sinx cosx - 1= 0`?

Answer 1

Reminder: sin 2a = 2sin a.cos a.

f(x) = sin 2x + 2sin x.cos x - 1 = 0
`2sin 2x = 1 -> sin 2x = 1/2 = sin (pi/6), and 1/2 = sin (5pi)/6`
On the trig unit circle:

a. `2x = pi/6 -> x = pi/12`

b. `2x = (5pi)/6 -> x = (5pi)/12.`

Check:
`x = pi/12 -> 2x = pi/6 -> sin 2x = 1/2` -> 1 - 1 = 0 Correct
`x = (5pi)/12 -> 2x = (5pi)/6 -> sin ((5pi)/6) = 1/2` -> 1 - 1 = 0 Correct