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

The general solution for sec(θ)+1/(cot(θ))=1 is No Solution for θ\in\mathbb{R}

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sec(θ)+1/(cot(θ))=1

sec (θ) + \frac{1}{cot ( θ )} = 1

No Solution for θ \in R

Solution steps

sec (θ) + \frac{1}{cot ( θ )} = 1

Subtract

1

from both sides

sec (θ) + \frac{1}{cot ( θ )} - 1 = 0

Simplify

sec (θ) + \frac{1}{cot ( θ )} - 1 : \frac{sec ( θ ) cot ( θ ) + 1 - cot ( θ )}{cot ( θ )}

\frac{sec ( θ ) cot ( θ ) + 1 - cot ( θ )}{cot ( θ )} = 0

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

sec (θ) cot (θ) + 1 - cot (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{5 π}{4} + 2 πn, θ - \frac{π}{4} = \frac{7 π}{4} + 2 πn

Solve

θ - \frac{π}{4} = \frac{5 π}{4} + 2 πn : θ = 2 πn + \frac{3 π}{2}

Solve

θ - \frac{π}{4} = \frac{7 π}{4} + 2 πn : θ = 2 πn + 2 π

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

Since the equation is undefined for:

2 πn + \frac{3 π}{2}, 2 πn + 2 π

No Solution for θ \in R

(1 + 2 sin (2 θ))^{\frac{1}{3}} = 0

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6 sin (3 x) = 5

x, 4 sin^{2} (x) = 5 - 4 cos (x)

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What is the general solution for sec(θ)+1/(cot(θ))=1 ?
The general solution for sec(θ)+1/(cot(θ))=1 is No Solution for θ\in\mathbb{R}