What is the general solution for 1+cot(θ)=csc(θ) ?

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

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1+cot(θ)=csc(θ)

1 + cot (θ) = csc (θ)

θ = 2 πn + \frac{π}{2}

Solution steps

1 + cot (θ) = csc (θ)

Subtract

csc (θ)

from both sides

1 + cot (θ) - csc (θ) = 0

Express with sin, cos

\frac{- 1 + cos ( θ ) + sin ( θ )}{sin ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 1 + cos (θ) + sin (θ) = 0

Rewrite using trig identities

- 1 + 2 sin (θ + \frac{π}{4}) = 0

Move

1

to the right side

2 sin (θ + \frac{π}{4}) = 1

Divide both sides by

2

sin (θ + \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (θ + \frac{π}{4}) = \frac{2}{2}

θ + \frac{π}{4} = \frac{π}{4} + 2 πn, θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

Solve

θ + \frac{π}{4} = \frac{π}{4} + 2 πn : θ = 2 πn

Solve

θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn : θ = 2 πn + \frac{π}{2}

θ = 2 πn, θ = 2 πn + \frac{π}{2}

Since the equation is undefined for:

2 πn

θ = 2 πn + \frac{π}{2}

tan (x) = cot (2 5^{\circ})

3 csc^{2} (5 x) = 4

3 sin^{2} (θ) + 2 cos (2 θ) = 1, 0 \leq θ \leq 2 π

2 cos^{2} (x) + 3 sin (- x) - 3 = 0

12 cos (\frac{x}{3}) = 0

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