PRE CALCULUS HELP PLEASEEEE !!!!?

Question

Solve this equation. Give a general formula for all the solutions using k to represent your arbitrary integer and using only angles between 0 and 2π.
1) tan theta= -sqrt3/3
(should be two different answers)

Solve this equation on the interval 0 θ < 2π. Round your answers to two decimal places. If there is no solution, enter NONE.
2) cos(2theta) + 6 sin^2 theta= 4
(should be 4 different answers)

For each of the following, find all solutions. The variable n denotes a generic integer. (For each, select all that apply.)

3) cot^2(x)-3=0
a)none of these
b)pi/3+2pi n
c)5pi/6 + pi n
d)pi/3 + pi n
e) pi/6 + pi n

Answer 1

3). cot²(x) - 3 = 0

Add 3 to each side.

cot²(x) = 3.

Take square root each side.

√[cot²(x)] = √3.

cot(x) = √3

Trigonometric table in cot(30°) = √3

cot(x) = cot(30°)

x = 30°= π/6

Therefore x = π/6 + nπ where n is an integer.

Option e is right choice.

Answer 2

1) tan(θ) = -√3/3

This expression it can be written as tan(θ) = -[√3/(√3)(√3)]

So, tan(θ) = -1/√3

Trigonometric table in tan(180 - 30) = -1/√3 or tan(360 - 30) = -1/√3

i.e. tan(150°) = -1/√3 or tan(330°) = -1/√3

tan(θ) = tan(150°) or tan(θ) = tan(330°)

θ = 150° = 5π/6 + nπ or 330°= 11π/6 + nπ

Therefore θ = 5π/6 + nπ or 11π/6 + nπ.

Answer 3

2) The trigonometric equation

The general solution of where n  is an integer.

or

The solutions outside the interval [0, 2π ) are

If n = 0 then

If n = 2 then

The solutions in the interval [0, 2π) are

The solutions in the interval [0, 2π) are π/3, 2π/3, 4π/3 and 5π/3.