What is the general solution for 3csc^2(θ)=1+2csc^2(θ) ?

The general solution for 3csc^2(θ)=1+2csc^2(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin

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3csc^2(θ)=1+2csc^2(θ)

3 csc^{2} (θ) = 1 + 2 csc^{2} (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

3 csc^{2} (θ) = 1 + 2 csc^{2} (θ)

Solve by substitution

3 csc^{2} (θ) = 1 + 2 csc^{2} (θ)

Let:

csc (θ) = u

3 u^{2} = 1 + 2 u^{2}

3 u^{2} = 1 + 2 u^{2} : u = 1, u = - 1

3 u^{2} = 1 + 2 u^{2}

Move

2 u^{2}

to the left side

3 u^{2} = 1 + 2 u^{2}

Subtract

2 u^{2}

from both sides

3 u^{2} - 2 u^{2} = 1 + 2 u^{2} - 2 u^{2}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = csc (θ)

csc (θ) = 1, csc (θ) = - 1

csc (θ) = 1, csc (θ) = - 1

csc (θ) = 1 : θ = \frac{π}{2} + 2 πn

csc (θ) = 1

General solutions for

csc (θ) = 1

csc (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

csc (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

csc (θ) = - 1

General solutions for

csc (θ) = - 1

csc (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

tan^{2} (x) = 2

0 = 2 cos (2 x)

4 cos^{4} (x) - 5 cos^{2} (x) + 1 = 0

3 sin (x) + 2 = sin (x)

sin (x) cos (x) - sin (x) = 0

What is the general solution for 3csc^2(θ)=1+2csc^2(θ) ?
The general solution for 3csc^2(θ)=1+2csc^2(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin