What is the general solution for cot(u)-csc(u)=(sin(u))/(1+cos(u)) ?

The general solution for cot(u)-csc(u)=(sin(u))/(1+cos(u)) is No Solution for u\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cot(u)-csc(u)=(sin(u))/(1+cos(u))

cot (u) - csc (u) = \frac{sin ( u )}{1 + cos ( u )}

No Solution for u \in R

Solution steps

cot (u) - csc (u) = \frac{sin ( u )}{1 + cos ( u )}

Subtract

\frac{sin ( u )}{1 + cos ( u )}

from both sides

cot (u) - csc (u) - \frac{sin ( u )}{1 + cos ( u )} = 0

Simplify

cot (u) - csc (u) - \frac{sin ( u )}{1 + cos ( u )} : \frac{cot ( u ) ( 1 + cos ( u ) ) - csc ( u ) ( 1 + cos ( u ) ) - sin ( u )}{1 + cos ( u )}

\frac{cot ( u ) ( 1 + cos ( u ) ) - csc ( u ) ( 1 + cos ( u ) ) - sin ( u )}{1 + cos ( u )} = 0

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

cot (u) (1 + cos (u)) - csc (u) (1 + cos (u)) - sin (u) = 0

Express with sin, cos

\frac{- 1 + cos ^{2} ( u ) - sin ^{2} ( u )}{sin ( u )} = 0

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

- 1 + cos^{2} (u) - sin^{2} (u) = 0

Rewrite using trig identities

- 1 + cos (2 u) = 0

Move

1

to the right side

cos (2 u) = 1

General solutions for

cos (2 u) = 1

2 u = 0 + 2 πn

Solve

2 u = 0 + 2 πn : u = πn

u = πn

Since the equation is undefined for:

πn

No Solution for u \in R

sin (x) = \frac{5}{3}

sin (θ) = \frac{3}{4}

2 cos (2 θ) - 1 = 0, 0 \leq θ \leq 2 π

6 cos (x) + 6 sin (x) tan (x) = 12, 0 \leq x \leq 2 π

6 ∣ tan (x) ∣ = 6

What is the general solution for cot(u)-csc(u)=(sin(u))/(1+cos(u)) ?
The general solution for cot(u)-csc(u)=(sin(u))/(1+cos(u)) is No Solution for u\in\mathbb{R}