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

The general solution for 2sin^2(2x)-sin(2x)-3=0 is x=(3pi)/4+pin

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2sin^2(2x)-sin(2x)-3=0

Solution

2 sin^{2} (2 x) - sin (2 x) - 3 = 0

Solution

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

2 sin^{2} (2 x) - sin (2 x) - 3 = 0

Solve by substitution

2 sin^{2} (2 x) - sin (2 x) - 3 = 0

Let:

sin (2 x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 \cdot 2 (- 3) = 5

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

Apply rule

- (- a) = a

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

(- 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 3 = 24

4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 5}{2 \cdot 2} : \frac{3}{2}

\frac{- ( - 1 ) + 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 5}{2 \cdot 2}

Add the numbers:

1 + 5 = 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

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

\frac{- ( - 1 ) - 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 5}{2 \cdot 2}

Subtract the numbers:

1 - 5 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (2 x)

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

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

sin (2 x) = \frac{3}{2} :

No Solution

sin (2 x) = \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (2 x) = - 1 : x = \frac{3 π}{4} + πn

sin (2 x) = - 1

General solutions for

sin (2 x) = - 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 x = \frac{3 π}{2} + 2 πn

2 x = \frac{3 π}{2} + 2 πn

Solve

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(2x)-sin(2x)-3=0 ?
The general solution for 2sin^2(2x)-sin(2x)-3=0 is x=(3pi)/4+pin