Absolute value equations with trig, go over my answers it they're correct thanks?

Question

Check my answers for these 6 questions please. Teacher didn't provide answer sheet. -_- 
-Solve absolute value equation in the interval [0,π] 
1) |sin(2x)|=√3 
Used double angle formula. I got x= π/2, π/3, 2π/3 
2) |3cotxcos²x|=cotx 
I got x= .96 QI and x=2.19 QII because cosx=+-(1/√3) 
3) |4sin²x|=3 
I got x=π/3 , 2π/3 
4) |2sin²x+1|=3sinx 
I got x=π/6,π/2,5π/6 
5) |√3tanx+2|=3 
I got x=π/6 
6) |tan²x-2|=1 
I got x=π/4 and x=.785 QI 
*Reminder solution has to be [0,π]

Answer 1

• 1). |sin(2x)| = √3.

sin(2x) = √3 or sin(2x) = - √3

2x = sin^-1 (√3) or 2x = sin^-1 (- √3)

⇒ x = (1/2)sin^-1 (√3) or x = (1/2)sin^-1 (- √3)

⇒ x = 0.506 or x = - 0.506.

Solution is x = 0.506 in the interval [0, π].

• 2). | 3 cot x cos² x| = cot x

3 cot x cos² x = cot x or 3 cot x cos² x = - cot x

3 cos² x = 1 or 3 cos² x = - 1

cos² x = 1/3 or cos² x = - 1/3

cos x = 1/√3 or cos x = - 1/√3

x = cos^{- 1} (1/√3) or x = cos^{- 1} (- 1/√3)

⇒ x = 0.96 or x = 2.19.               (By using calculator)

Solution is x = 0.96 or x = 2.19 in the interval [0, π].

• 3). |4sin² x| = 3.

4sin² x = 3 or 4sin² x = - 3

sin² x = 3/4 or sin² x = - 3/4

sin x = √3/2 or sin x = - √3/2

sin x = sin(π/3) and sin x = sin(π/3) = sin(π - π/3) = sin(2π/3) in the interval [0, π].

⇒ x = π/3 and x = 2π/3.

Solution is x = π/3 and x = 2π/3 in the interval [0, π].

• 4). |2sin² x + 1| = 3sin x.

2sin² x + 1 = 3sin x or 2sin² x + 1 = - 3sin x

2sin² x - 3sin x + 1 = 0 or 2sin² x + 3sin x + 1 = 0

2sin² x - 2sin x - sin x + 1 = 0 or 2sin² x + 2sin x + sin x + 1 = 0

2sin x (sin x - 1) - 1(sin x - 1) = 0 or 2sin x(sin x + 1) + 1(sin x + 1) = 0

(sin x - 1)(2sin x - 1) = 0 or (sin x + 1)(2sin x + 1) = 0

Solve : (sin x - 1)(2sin x - 1) = 0.

sin x - 1 = 0 and 2sin x - 1 = 0

⇒ sin x = 1 and sin x = 1/2

sin x = sin(π/2) and sin x = sin(π/6)

⇒ x = π/2 and x = π/6.

Solve : (sin x + 1)(2sin x + 1) = 0.

sin x + 1 = 0 and 2sin x + 1 = 0

⇒ sin x = - 1 and sin x = - 1/2

sin x = sin(- π/2) and sin x = sin(- π/6) = sin(π- π/6) = sin(5π/6)

⇒ x = - π/2 and x = 5π/6.

Solutions are  x = π/2, x = π/6, and x = 5π/6 in the interval [0, π].

Answer 2

Contd....

• 5). | |√3 tan x + 2| = 3.

√3 tan x + 2 = 3 or √3 tan x + 2 = - 3

√3 tan x = 1 or √3 tan x = - 5

tan x = 1/√3 or tan x = - 5/√3

⇒ x = tan^{- 1}(tan(π/6)) or x = tan^{- 1}(- 5/√3).

Solution is x = π/6 in the interval [0, π].

• 6). |tan² x - 2| = 1.

tan² x - 2 = 1 or tan² x - 2 = - 1

tan² x = 3 or tan² x = 1

tan x = √3 or tan x = 1

x = tan^{- 1} (√3 ) or x = tan^{- 1}(1)

⇒ x = π/3 and x = π/4.

Solutions are  x = π/3 and x = π/4 in the interval [0, π].