What is the general solution for 3sin^2(x)+4sin(x)+1=0 ?

The general solution for 3sin^2(x)+4sin(x)+1=0 is x=-0.33983…+2pin,x=pi+0.33983…+2pin,x=(3pi)/2+2pin

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3sin^2(x)+4sin(x)+1=0

Solution

3 sin^{2} (x) + 4 sin (x) + 1 = 0

Solution

x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 199.47122 \dots^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin^{2} (x) + 4 sin (x) + 1 = 0

Solve by substitution

3 sin^{2} (x) + 4 sin (x) + 1 = 0

Let:

sin (x) = u

3 u^{2} + 4 u + 1 = 0

3 u^{2} + 4 u + 1 = 0 : u = - \frac{1}{3}, u = - 1

3 u^{2} + 4 u + 1 = 0

Solve with the quadratic formula

3 u^{2} + 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 3, b = 4, c = 1

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

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

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

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

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

u = \frac{- 4 + 2}{2 \cdot 3} : - \frac{1}{3}

\frac{- 4 + 2}{2 \cdot 3}

Add/Subtract the numbers:

- 4 + 2 = - 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{6}

Apply the fraction rule:

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

= - \frac{2}{6}

Cancel the common factor:

2

= - \frac{1}{3}

u = \frac{- 4 - 2}{2 \cdot 3} : - 1

\frac{- 4 - 2}{2 \cdot 3}

Subtract the numbers:

- 4 - 2 = - 6

= \frac{- 6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{6}

Apply the fraction rule:

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

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - \frac{1}{3}, u = - 1

Substitute back

u = sin (x)

sin (x) = - \frac{1}{3}, sin (x) = - 1

sin (x) = - \frac{1}{3}, sin (x) = - 1

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

sin (x) = - \frac{1}{3}

Apply trig inverse properties

sin (x) = - \frac{1}{3}

General solutions for

sin (x) = - \frac{1}{3}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

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

sin (x) = - 1

General solutions for

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn, x = \frac{3 π}{2} + 2 πn

Show solutions in decimal form

x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn, x = \frac{3 π}{2} + 2 πn

Popular Examples

cos(x)= 9/14

cos (x) = \frac{9}{14}

sin(θ)= 7/9

sin (θ) = \frac{7}{9}

cos(θ)=0.97617

cos (θ) = 0.97617

cos(a)=0,999^{99}x,(sin(a))/(cos(a))x

cos (a) = 0, 99 9^{99} x, \frac{sin ( a )}{cos ( a )} x

sin(x/2+pi)= 3/(2sqrt(3))

sin (\frac{x}{2} + π) = \frac{3}{2 3}

Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(x)+4sin(x)+1=0 ?
The general solution for 3sin^2(x)+4sin(x)+1=0 is x=-0.33983…+2pin,x=pi+0.33983…+2pin,x=(3pi)/2+2pin