What is the general solution for 5sin^2(θ)-11sin(θ)+2=0 ?

The general solution for 5sin^2(θ)-11sin(θ)+2=0 is θ=0.20135…+2pin,θ=pi-0.20135…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5sin^2(θ)-11sin(θ)+2=0

Solution

5 sin^{2} (θ) - 11 sin (θ) + 2 = 0

Solution

θ = 0.20135 \dots + 2 πn, θ = π - 0.20135 \dots + 2 πn

Degrees

θ = 11.53695 \dots^{\circ} + 36 0^{\circ} n, θ = 168.46304 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin^{2} (θ) - 11 sin (θ) + 2 = 0

Solve by substitution

5 sin^{2} (θ) - 11 sin (θ) + 2 = 0

Let:

sin (θ) = u

5 u^{2} - 11 u + 2 = 0

5 u^{2} - 11 u + 2 = 0 : u = 2, u = \frac{1}{5}

5 u^{2} - 11 u + 2 = 0

Solve with the quadratic formula

5 u^{2} - 11 u + 2 = 0

Quadratic Equation Formula:

For

a = 5, b = - 11, c = 2

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 5 \cdot 2}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 5 \cdot 2}{2 \cdot 5}

(- 11)^{2} - 4 \cdot 5 \cdot 2 = 9

(- 11)^{2} - 4 \cdot 5 \cdot 2

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 11)^{2} = 1 1^{2}

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

Multiply the numbers:

4 \cdot 5 \cdot 2 = 40

= 1 1^{2} - 40

1 1^{2} = 121

= 121 - 40

Subtract the numbers:

121 - 40 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 11 ) \pm 9}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 9}{2 \cdot 5}, u_{2} = \frac{- ( - 11 ) - 9}{2 \cdot 5}

u = \frac{- ( - 11 ) + 9}{2 \cdot 5} : 2

\frac{- ( - 11 ) + 9}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{11 + 9}{2 \cdot 5}

Add the numbers:

11 + 9 = 20

= \frac{20}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{20}{10}

Divide the numbers:

\frac{20}{10} = 2

= 2

u = \frac{- ( - 11 ) - 9}{2 \cdot 5} : \frac{1}{5}

\frac{- ( - 11 ) - 9}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{11 - 9}{2 \cdot 5}

Subtract the numbers:

11 - 9 = 2

= \frac{2}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{2}{10}

Cancel the common factor:

2

= \frac{1}{5}

The solutions to the quadratic equation are:

u = 2, u = \frac{1}{5}

Substitute back

u = sin (θ)

sin (θ) = 2, sin (θ) = \frac{1}{5}

sin (θ) = 2, sin (θ) = \frac{1}{5}

sin (θ) = 2 :

No Solution

sin (θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

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

sin (θ) = \frac{1}{5}

Apply trig inverse properties

sin (θ) = \frac{1}{5}

General solutions for

sin (θ) = \frac{1}{5}

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

θ = arcsin (\frac{1}{5}) + 2 πn, θ = π - arcsin (\frac{1}{5}) + 2 πn

θ = arcsin (\frac{1}{5}) + 2 πn, θ = π - arcsin (\frac{1}{5}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1}{5}) + 2 πn, θ = π - arcsin (\frac{1}{5}) + 2 πn

Show solutions in decimal form

θ = 0.20135 \dots + 2 πn, θ = π - 0.20135 \dots + 2 πn

Popular Examples

tan^2(θ)+3sec(θ)=-3

tan^{2} (θ) + 3 sec (θ) = - 3

cos^2(x)-sin^2(x)=1-sin^2(x)

cos^{2} (x) - sin^{2} (x) = 1 - sin^{2} (x)

sin(2x)cos(x)+sin(x)=0

sin (2 x) cos (x) + sin (x) = 0

cos(2x)+cos(x)=1+sin(2x)-sin(x)

cos (2 x) + cos (x) = 1 + sin (2 x) - sin (x)

-sin(θ)+2sin^2(θ)=0

- sin (θ) + 2 sin^{2} (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 5sin^2(θ)-11sin(θ)+2=0 ?
The general solution for 5sin^2(θ)-11sin(θ)+2=0 is θ=0.20135…+2pin,θ=pi-0.20135…+2pin