What is the general solution for 3sin(t)=1+sin(t) ?

The general solution for 3sin(t)=1+sin(t) is t= pi/6+2pin,t=(5pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3sin(t)=1+sin(t)

3 sin (t) = 1 + sin (t)

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Degrees

t = 3 0^{\circ} + 36 0^{\circ} n, t = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (t) = 1 + sin (t)

Solve by substitution

3 sin (t) = 1 + sin (t)

Let:

sin (t) = u

3 u = 1 + u

3 u = 1 + u : u = \frac{1}{2}

3 u = 1 + u

Move

u

to the left side

3 u = 1 + u

Subtract

u

from both sides

3 u - u = 1 + u - u

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = sin (t)

sin (t) = \frac{1}{2}

sin (t) = \frac{1}{2}

sin (t) = \frac{1}{2} : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

sin (t) = \frac{1}{2}

General solutions for

sin (t) = \frac{1}{2}

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}

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Combine all the solutions

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

sin^{2} (t) sin (t) = 0

cos (x) = \frac{100}{130}

sec^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

2 cos^{2} (θ) - 3 cos (θ) = 0, \frac{3 π}{2} < θ < 2 π

cos (θ) = \frac{9}{17}

What is the general solution for 3sin(t)=1+sin(t) ?
The general solution for 3sin(t)=1+sin(t) is t= pi/6+2pin,t=(5pi)/6+2pin