What is the general solution for 2sin(2θ)-3-6/(sin(2θ)-1)=0 ?

The general solution for 2sin(2θ)-3-6/(sin(2θ)-1)=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin

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2sin(2θ)-3-6/(sin(2θ)-1)=0

Solution

2 sin (2 θ) - 3 - \frac{6}{sin ( 2 θ ) - 1} = 0

Solution

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

Degrees

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

Solution steps

2 sin (2 θ) - 3 - \frac{6}{sin ( 2 θ ) - 1} = 0

Solve by substitution

2 sin (2 θ) - 3 - \frac{6}{sin ( 2 θ ) - 1} = 0

Let:

sin (2 θ) = u

2 u - 3 - \frac{6}{u - 1} = 0

2 u - 3 - \frac{6}{u - 1} = 0 : u = 3, u = - \frac{1}{2}

2 u - 3 - \frac{6}{u - 1} = 0

Multiply both sides by

u - 1

2 u - 3 - \frac{6}{u - 1} = 0

Multiply both sides by

u - 1

2 u (u - 1) - 3 (u - 1) - \frac{6}{u - 1} (u - 1) = 0 \cdot (u - 1)

Simplify

2 u (u - 1) - 3 (u - 1) - \frac{6}{u - 1} (u - 1) = 0 \cdot (u - 1)

Simplify

- \frac{6}{u - 1} (u - 1) : - 6

- \frac{6}{u - 1} (u - 1)

Multiply fractions:

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

= - \frac{6 ( u - 1 )}{u - 1}

Cancel the common factor:

u - 1

= - 6

Simplify

0 \cdot (u - 1) : 0

0 \cdot (u - 1)

Apply rule

0 \cdot a = 0

= 0

2 u (u - 1) - 3 (u - 1) - 6 = 0

2 u (u - 1) - 3 (u - 1) - 6 = 0

2 u (u - 1) - 3 (u - 1) - 6 = 0

Solve

2 u (u - 1) - 3 (u - 1) - 6 = 0 : u = 3, u = - \frac{1}{2}

2 u (u - 1) - 3 (u - 1) - 6 = 0

Expand

2 u (u - 1) - 3 (u - 1) - 6 : 2 u^{2} - 5 u - 3

2 u (u - 1) - 3 (u - 1) - 6

Expand

2 u (u - 1) : 2 u^{2} - 2 u

2 u (u - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2 u, b = u, c = 1

= 2 uu - 2 u \cdot 1

= 2 uu - 2 \cdot 1 \cdot u

Simplify

2 uu - 2 \cdot 1 \cdot u : 2 u^{2} - 2 u

2 uu - 2 \cdot 1 \cdot u

2 uu = 2 u^{2}

2 uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2 u^{2}

2 \cdot 1 \cdot u = 2 u

2 \cdot 1 \cdot u

Multiply the numbers:

2 \cdot 1 = 2

= 2 u

= 2 u^{2} - 2 u

= 2 u^{2} - 2 u

= 2 u^{2} - 2 u - 3 (u - 1) - 6

Expand

- 3 (u - 1) : - 3 u + 3

- 3 (u - 1)

Apply the distributive law:

a (b - c) = ab - a c

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

= - 3 u - (- 3) \cdot 1

Apply minus-plus rules

- (- a) = a

= - 3 u + 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= - 3 u + 3

= 2 u^{2} - 2 u - 3 u + 3 - 6

Simplify

2 u^{2} - 2 u - 3 u + 3 - 6 : 2 u^{2} - 5 u - 3

2 u^{2} - 2 u - 3 u + 3 - 6

Add similar elements:

- 2 u - 3 u = - 5 u

= 2 u^{2} - 5 u + 3 - 6

Add/Subtract the numbers:

3 - 6 = - 3

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

= 2 u^{2} - 5 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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 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} : 3

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

Apply rule

- (- a) = a

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

Add the numbers:

5 + 7 = 12

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

5 - 7 = - 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 = 3, u = - \frac{1}{2}

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

Verify Solutions

Find undefined (singularity) points:

u = 1

Take the denominator(s) of

2 u - 3 - \frac{6}{u - 1}

and compare to zero

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

The following points are undefined

u = 1

Combine undefined points with solutions:

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

Substitute back

u = sin (2 θ)

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

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

sin (2 θ) = 3 :

No Solution

sin (2 θ) = 3

- 1 \leq sin (x) \leq 1

No Solution

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{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(2θ)-3-6/(sin(2θ)-1)=0 ?
The general solution for 2sin(2θ)-3-6/(sin(2θ)-1)=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin