How do you solve $sqrt(3)sin2x+cos2x=0$ for 0 to 2pi?

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Answer 1 · 2015-06-15T17:55:31

Solve $sqrt3.sin 2x + cos 2x = 0$

$sin 2x + (1/sqrt3)cos 2x = 0$ .

Replace $tan (pi/6) = (1/sqrt3)$ by $(sin (pi/6)/cos (pi/6))$

$sin 2x.cos (pi/6) + sin (pi/6).cos 2x = 0$

$sin (2x + pi/6) = 0$

a. $(2x + pi/6) = 0$ -> $2x = - pi/6 -> x = - pi/12$

b. $(2x + pi/6) = pi$ -> $2x = pi - pi/6 = (5pi)/6$ ->

-> $x = 5pi/12$

How do you solve sqrt(3)sin2x+cos2x=0sqrt(3)sin2x+cos2x=0 for 0 to 2pi?