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

1 Answer

May 14, 2015

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

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