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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^2(θ)=2sin(θ)+3

Solution

sin^{2} (θ) = 2 sin (θ) + 3

Solution

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (θ) = 2 sin (θ) + 3

Solve by substitution

sin^{2} (θ) = 2 sin (θ) + 3

Let:

sin (θ) = u

u^{2} = 2 u + 3

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

u^{2} = 2 u + 3

Move

3

to the left side

u^{2} = 2 u + 3

Subtract

3

from both sides

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

Simplify

u^{2} - 3 = 2 u

u^{2} - 3 = 2 u

Move

2 u

to the left side

u^{2} - 3 = 2 u

Subtract

2 u

from both sides

u^{2} - 3 - 2 u = 2 u - 2 u

Simplify

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 3

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

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

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

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

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 3

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

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

\frac{- ( - 2 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 4}{2 \cdot 1}

Add the numbers:

2 + 4 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

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

\frac{- ( - 2 ) - 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 4}{2 \cdot 1}

Subtract the numbers:

2 - 4 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 3, u = - 1

Substitute back

u = sin (θ)

sin (θ) = 3, sin (θ) = - 1

sin (θ) = 3, sin (θ) = - 1

sin (θ) = 3 :

No Solution

sin (θ) = 3

- 1 \leq sin (x) \leq 1

No Solution

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

sin (θ) = - 1

General solutions for

sin (θ) = - 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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

Popular Examples

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

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

sec(x)=sqrt(1-tan(x))

sec (x) = 1 - tan (x)

-cos(x)=cos(x)

- cos (x) = cos (x)

sqrt(2)cos(θ)-1=0

2 cos (θ) - 1 = 0

6sin^2(x)-7sin(x)+2=0

6 sin^{2} (x) - 7 sin (x) + 2 = 0

Frequently Asked Questions (FAQ)

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