What is the general solution for 2sin^2(t)+3sin(t)+1=0 ?

The general solution for 2sin^2(t)+3sin(t)+1=0 is t=(7pi)/6+2pin,t=(11pi)/6+2pin,t=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(t)+3sin(t)+1=0

Solution

2 sin^{2} (t) + 3 sin (t) + 1 = 0

Solution

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn, t = \frac{3 π}{2} + 2 πn

Degrees

t = 21 0^{\circ} + 36 0^{\circ} n, t = 33 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (t) + 3 sin (t) + 1 = 0

Solve by substitution

2 sin^{2} (t) + 3 sin (t) + 1 = 0

Let:

sin (t) = u

2 u^{2} + 3 u + 1 = 0

2 u^{2} + 3 u + 1 = 0 : u = - \frac{1}{2}, u = - 1

2 u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

3^{2} - 4 \cdot 2 \cdot 1 = 1

3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 1}{2 \cdot 2}, u_{2} = \frac{- 3 - 1}{2 \cdot 2}

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

\frac{- 3 + 1}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 1 = - 2

= \frac{- 2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 3 - 1}{2 \cdot 2} : - 1

\frac{- 3 - 1}{2 \cdot 2}

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (t)

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

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

sin (t) = - \frac{1}{2} : t = \frac{7 π}{6} + 2 πn, t = \frac{11 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{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

sin (t) = - 1 : t = \frac{3 π}{2} + 2 πn

sin (t) = - 1

General solutions for

sin (t) = - 1

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{3 π}{2} + 2 πn

t = \frac{3 π}{2} + 2 πn

Combine all the solutions

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn, t = \frac{3 π}{2} + 2 πn

Popular Examples

cos(x)=-0.6060400000000…

cos (x) = - 0.6060400000000 \dots

sin(x)cot(x)=0.5

sin (x) cot (x) = 0.5

sin(2x)tan(2x+2)=sin(3x-1)tan(2x+2)

sin (2 x) tan (2 x + 2) = sin (3 x - 1) tan (2 x + 2)

2sin^2(θ)-7sin(θ)-4=0

2 sin^{2} (θ) - 7 sin (θ) - 4 = 0

2tan(x)-2sqrt(3)=0

2 tan (x) - 23 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(t)+3sin(t)+1=0 ?
The general solution for 2sin^2(t)+3sin(t)+1=0 is t=(7pi)/6+2pin,t=(11pi)/6+2pin,t=(3pi)/2+2pin