Trigonometric Equation?

Question

Hey, how would you do this?

Solve for x, 0 ≤ x ≤ 360°, (sin^2)2x+5sin2x-2 = 4sin2x

Answer 1

sin²(2x) + sin(2x) - 2 = 0
(sin 2x + 2).(sin 2x - 1) = 0
sin 2x = 1 (is acceptable answer)
2x = 90° , 450°
x = 45° , 225°

Answer 2

let sin 2x=y then y^2+5y-2=4y
y^2+(5-4)y-2=0
y^2+y-2=0 two ways1:delta q 2:analyzis
y^2+y-2=(y-1).(y+2)=0 y=1or y=-2 the second is invalid because -1 x=45degree for 0

Answer 3

The person above me is correct except since your in Trig the answer would probably be in radians instead of degrees so instead of x = 45degrees it would be x=pi

Basically just subtract the 4sin2x from both sides giving you sin^2(2x) +sin(2x)-2=0 (the 5sin(2x) - 4sin(2x) gives sin(2x)).

From there you just factor and solve both.

Answer 4

Let y = sin2x.
y^2+y-2 = 0
(y+2)(y-1) = 0
y = -2 (discarded. Why?) or 1
sin2x = 1
2x = 90+360n
n = 0, x = 45°
n = 1, x = 225°

Answer 5

First get all the sines to one side.

(sin^2)2x + sin2x - 2 = 0

Factor the sines as you would a regular polynomial:

(sin2x+2)(sin2x-1) = 0

sin(2x) = 1

x= 45

sin2x = -2

There are no sine values that can equal -2, so throw that out.

I might be rusty in my trig.