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

The general solution for 2sin^2(θ)+5sin(θ)-3=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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2sin^2(θ)+5sin(θ)-3=0

Solution

2 sin^{2} (θ) + 5 sin (θ) - 3 = 0

Solution

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

Degrees

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

Solution steps

2 sin^{2} (θ) + 5 sin (θ) - 3 = 0

Solve by substitution

2 sin^{2} (θ) + 5 sin (θ) - 3 = 0

Let:

sin (θ) = u

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

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

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 ( - 3 )}{2 \cdot 2}

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

5^{2} - 4 \cdot 2 (- 3) = 7

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

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 5 + 7}{2 \cdot 2}, u_{2} = \frac{- 5 - 7}{2 \cdot 2}

u = \frac{- 5 + 7}{2 \cdot 2} : \frac{1}{2}

\frac{- 5 + 7}{2 \cdot 2}

Add/Subtract the numbers:

- 5 + 7 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 5 - 7}{2 \cdot 2} : - 3

\frac{- 5 - 7}{2 \cdot 2}

Subtract the numbers:

- 5 - 7 = - 12

= \frac{- 12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 12}{4}

Apply the fraction rule:

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

= - \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= - 3

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - 3

sin (θ) = \frac{1}{2}, sin (θ) = - 3

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

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \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}

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

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

sin (θ) = - 3 :

No Solution

sin (θ) = - 3

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

Popular Examples

sin(2x)-sin(x)=0,-pi<= x<= pi

sin (2 x) - sin (x) = 0, - π \leq x \leq π

2(tan(x)+3)=5+tan(x)

2 (tan (x) + 3) = 5 + tan (x)

3cos(4θ)=-2

3 cos (4 θ) = - 2

solvefor x,sin(x)=sin(y)solve for

x, sin (x) = sin (y)

cos^2(x)=((5k-3))/3

cos^{2} (x) = \frac{( 5 k - 3 )}{3}

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)+5sin(θ)-3=0 ?
The general solution for 2sin^2(θ)+5sin(θ)-3=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin