What is the general solution for 2sin^2(2θ)=1+sin(2θ) ?

The general solution for 2sin^2(2θ)=1+sin(2θ) is θ= pi/4+pin,θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin

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2sin^2(2θ)=1+sin(2θ)

Solution

2 sin^{2} (2 θ) = 1 + sin (2 θ)

Solution

θ = \frac{π}{4} + πn, θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 10 5^{\circ} + 18 0^{\circ} n, θ = 16 5^{\circ} + 18 0^{\circ} n

Solution steps

2 sin^{2} (2 θ) = 1 + sin (2 θ)

Solve by substitution

2 sin^{2} (2 θ) = 1 + sin (2 θ)

Let:

sin (2 θ) = u

2 u^{2} = 1 + u

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

2 u^{2} = 1 + u

Move

u

to the left side

2 u^{2} = 1 + u

Subtract

u

from both sides

2 u^{2} - u = 1 + u - u

Simplify

2 u^{2} - u = 1

2 u^{2} - u = 1

Move

1

to the left side

2 u^{2} - u = 1

Subtract

1

from both sides

2 u^{2} - u - 1 = 1 - 1

Simplify

2 u^{2} - u - 1 = 0

2 u^{2} - u - 1 = 0

Solve with the quadratic formula

2 u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 3}{2 \cdot 2}

Add the numbers:

1 + 3 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

\frac{- ( - 1 ) - 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 3}{2 \cdot 2}

Subtract the numbers:

1 - 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (2 θ)

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

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

sin (2 θ) = 1 : θ = \frac{π}{4} + πn

sin (2 θ) = 1

General solutions for

sin (2 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}

2 θ = \frac{π}{2} + 2 πn

2 θ = \frac{π}{2} + 2 πn

Solve

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

sin (2 θ) = - \frac{1}{2} : θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

sin (2 θ) = - \frac{1}{2}

General solutions for

sin (2 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}

2 θ = \frac{7 π}{6} + 2 πn, 2 θ = \frac{11 π}{6} + 2 πn

2 θ = \frac{7 π}{6} + 2 πn, 2 θ = \frac{11 π}{6} + 2 πn

Solve

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

2 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π}{12} + πn

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{7 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

Solve

2 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{12} + πn

2 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π}{12} + πn

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{11 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(2θ)=1+sin(2θ) ?
The general solution for 2sin^2(2θ)=1+sin(2θ) is θ= pi/4+pin,θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin