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

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Answer 1 · 2015-05-14T00:28:10

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

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