What is the general solution for csc(θ)+2.402=0 ?

The general solution for csc(θ)+2.402=0 is θ=-0.42939…+2pin,θ=pi+0.42939…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

csc(θ)+2.402=0

csc (θ) + 2.402 = 0

θ = - 0.42939 \dots + 2 πn, θ = π + 0.42939 \dots + 2 πn

Solution steps

csc (θ) + 2.402 = 0

Move

2.402

to the right side

csc (θ) = - 2.402

Apply trig inverse properties

θ = arccsc (- 2.402) + 2 πn, θ = π + arccsc (2.402) + 2 πn

Show solutions in decimal form

θ = - 0.42939 \dots + 2 πn, θ = π + 0.42939 \dots + 2 πn

tan^{2} (β) = 1

6 sin (C) + 8 = 0

cos (2 t) = - sin (t)

sin^{2} (x) - 5 cos (x) - 5 = 0

\frac{1 + tan ^{2} ( 27 \cdot θ )}{1 - tan ^{2} ( 27 \cdot θ )} = 2

What is the general solution for csc(θ)+2.402=0 ?
The general solution for csc(θ)+2.402=0 is θ=-0.42939…+2pin,θ=pi+0.42939…+2pin