What is the general solution for 2cot(θ)+csc^2(θ)=0 ?

The general solution for 2cot(θ)+csc^2(θ)=0 is θ=(3pi)/4+pin

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2cot(θ)+csc^2(θ)=0

2 cot (θ) + csc^{2} (θ) = 0

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

2 cot (θ) + csc^{2} (θ) = 0

Rewrite using trig identities

csc^{2} (θ) + 2 cot (θ)

Use the Pythagorean identity:

csc^{2} (x) = 1 + cot^{2} (x)

= 1 + cot^{2} (θ) + 2 cot (θ)

1 + cot^{2} (θ) + 2 cot (θ) = 0

Solve by substitution

1 + cot^{2} (θ) + 2 cot (θ) = 0

Let:

cot (θ) = u

1 + u^{2} + 2 u = 0

1 + u^{2} + 2 u = 0 : u = - 1

1 + u^{2} + 2 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + 2 u + 1 = 0

Solve with the quadratic formula

u^{2} + 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 1

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 1 = 0

2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

u = \frac{- 2}{2 \cdot 1}

\frac{- 2}{2 \cdot 1} = - 1

\frac{- 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = cot (θ)

cot (θ) = - 1

cot (θ) = - 1

cot (θ) = - 1 : θ = \frac{3 π}{4} + πn

cot (θ) = - 1

General solutions for

cot (θ) = - 1

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = \frac{3 π}{4} + πn

csc (x) cot (x) = \frac{2}{3}

2 + 2 cos (x) = 0

4 tan^{2} (x) - 8 tan (x) + 3 = 0

sin (t) - sin (2 t) = 0

sin (\frac{3 x}{2}) = 0

What is the general solution for 2cot(θ)+csc^2(θ)=0 ?
The general solution for 2cot(θ)+csc^2(θ)=0 is θ=(3pi)/4+pin