What is the general solution for 2sin^2(x)+5sin(x)+3=0 ?

The general solution for 2sin^2(x)+5sin(x)+3=0 is x=(3pi)/2+2pin

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Popular >

2sin^2(x)+5sin(x)+3=0

Solution

2 sin^{2} (x) + 5 sin (x) + 3 = 0

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (x) + 5 sin (x) + 3 = 0

Solve by substitution

2 sin^{2} (x) + 5 sin (x) + 3 = 0

Let:

sin (x) = u

2 u^{2} + 5 u + 3 = 0

2 u^{2} + 5 u + 3 = 0 : u = - 1, u = - \frac{3}{2}

2 u^{2} + 5 u + 3 = 0

Solve with the quadratic formula

2 u^{2} + 5 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = 5, c = 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 3 = 1

5^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 5 + 1}{2 \cdot 2}, u_{2} = \frac{- 5 - 1}{2 \cdot 2}

u = \frac{- 5 + 1}{2 \cdot 2} : - 1

\frac{- 5 + 1}{2 \cdot 2}

Add/Subtract the numbers:

- 5 + 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

u = \frac{- 5 - 1}{2 \cdot 2} : - \frac{3}{2}

\frac{- 5 - 1}{2 \cdot 2}

Subtract the numbers:

- 5 - 1 = - 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{4}

Apply the fraction rule:

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

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = - \frac{3}{2}

sin (x) = - 1, sin (x) = - \frac{3}{2}

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

sin (x) = - \frac{3}{2} :

No Solution

sin (x) = - \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Popular Examples

cos(A)= 4/5

cos (A) = \frac{4}{5}

cos(2x+15)=sin(x/2-15)

cos (2 x + 15) = sin (\frac{x}{2} - 1 5^{\circ})

arcsin(x-2)= pi/6

arcsin (x - 2) = \frac{π}{6}

9.8sin(x)-1.96cos(x)=8.54

9.8 sin (x) - 1.96 cos (x) = 8.54

4cos(θ)-2sqrt(3)=0

4 cos (θ) - 23 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)+5sin(x)+3=0 ?
The general solution for 2sin^2(x)+5sin(x)+3=0 is x=(3pi)/2+2pin