What is the general solution for 1+sin(θ)=1+sin(-θ) ?

The general solution for 1+sin(θ)=1+sin(-θ) is θ=2pin,θ=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1+sin(θ)=1+sin(-θ)

1 + sin (θ) = 1 + sin (- θ)

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + sin (θ) = 1 + sin (- θ)

Rewrite using trig identities

1 + sin (θ) = 1 + sin (- θ)

1 + sin (θ) = 1 - sin (θ)

1 + sin (θ) = 1 - sin (θ)

Subtract

1 - sin (θ)

from both sides

2 sin (θ) = 0

Divide both sides by

2

2 sin (θ) = 0

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{0}{2}

Simplify

sin (θ) = 0

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

cos (2 x) = 2 + 3 sin (x), \frac{π}{2} \leq x \leq \frac{3 π}{2}

tan (θ) = (\frac{3}{8})

29 = \frac{1}{2} (5.6) (7.1) sin (x)

\frac{10.4}{sin ( x )} = \frac{12}{sin ( 6 2 ^{\circ} )}

csc (θ) = - 2, 0^{\circ} \leq θ \leq 36 0^{\circ}

What is the general solution for 1+sin(θ)=1+sin(-θ) ?
The general solution for 1+sin(θ)=1+sin(-θ) is θ=2pin,θ=pi+2pin