HELP!! precalc: a graphical approach to precalculus

Question

solve the equations forsolutions in the interval [0, 2π):
sin^2(x)=1

2tanx-1=0

5cot^2(x)+3cotx=2

Answer 1

Equation : sin²(x) = 1 ⇒ sin(x) = ± 1.

sin(x) = 1 and sin(x) = - 1

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

General solution : If sin(θ) = sin(∝) then θ = nπ + (-1)ⁿ∝, where n is integer.

If ∝ = π/2 then x = nπ + (-1)ⁿ(π/2).

If ∝ = -π/2 then x = nπ + (-1)ⁿ(-π/2) ⇒ x = nπ - (-1)ⁿ(π/2).

If n = 0 then x = (0)π + (-1)⁰(π/2) = π/2.

If n = 1 then x = (1)π + (-1)¹(π/2) = 3π/2.

The solutions in the interval in (0, 2π] are π/2 and 3π/2.

 

Equation : 2 tan(x) - 1 = 0 ⇒ tan(x) = 1/2.

tan(x) = tan(26.565^o)

General solution : If tan(θ) = tan(∝) then θ = 180^on + ∝, where n is integer.

If ∝ = 26.565^o then x = 180^on + 26.565.

If n = 0 then x = 180^o(0) + 26.565^o = 26.565^o.

If n = 1 then x = 180^o(1) + 26.565^o = 206.565^o.

The solutions in the interval in (0, 2π] are 26.565^o and 206.565^o.

 

Equation : 5 cot²(x) + 3 cot(x) = 2.

5cot²(x) + 3cot(x) - 2 = 0

5cot²(x) + 5cot(x) - 2cot(x) - 2 = 0

5cot(x) * [cot(x) + 1] -2[cot(x) + 1] = 0

[5cot(x) - 2] * [cot(x) + 1] = 0

cot(x) = 2/5 and cot(x) = - 1

cot(x) = cot(68.199^o) and cot(x) = - 45^o

General solution : If cot(θ) = cot(∝) then θ = 180^on + ∝, where n is integer.

If ∝ = 68.199^o then x = 180^on + 68.199^o.

If n = 0 then x = 180(0) + 68.199^o = 68.199^o.

If n = 0 then x = 180(1) + 68.199^o = 248.199^o.

If ∝ = -45^o then x = 180^on - 45^o.

If n = 1 then x = 180(1) - 45^o = 135^o.

If n = 1 then x = 180(2) - 45^o = 315^o.

The solutions in the interval in (0, 2π] are 68.199^o, 248.199^o, 135^o and 315^o.