What is the general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 ?

Q: What is the general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 ?

The general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 is

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2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360

Solution

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

Solution

β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}, β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

Radians

β = \frac{π}{6}, β = \frac{5 π}{6}, β = \frac{7 π}{6}, β = \frac{11 π}{6}, β = - \frac{5 π}{6}, β = - \frac{π}{6}, β = - \frac{11 π}{6}, β = - \frac{7 π}{6}, β = \frac{π}{2}, β = \frac{3 π}{2}, β = - \frac{π}{2}, β = - \frac{3 π}{2}

Solution steps

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

Solve by substitution

2 cos^{2} (2 β) + cos (2 β) - 1 = 0

Let:

cos (2 β) = u

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

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

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

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

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

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

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Subtract the numbers:

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

Substitute back

u = cos (2 β)

cos (2 β) = \frac{1}{2}, cos (2 β) = - 1

cos (2 β) = \frac{1}{2}, cos (2 β) = - 1

cos (2 β) = \frac{1}{2}, - 36 0^{\circ} \leq β \leq 36 0^{\circ} : β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}

cos (2 β) = \frac{1}{2}, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

General solutions for

cos (2 β) = \frac{1}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 β = 6 0^{\circ} + 36 0^{\circ} n, 2 β = 30 0^{\circ} + 36 0^{\circ} n

2 β = 6 0^{\circ} + 36 0^{\circ} n, 2 β = 30 0^{\circ} + 36 0^{\circ} n

Solve

2 β = 6 0^{\circ} + 36 0^{\circ} n : β = 3 0^{\circ} + 18 0^{\circ} n

2 β = 6 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 β = 6 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 β}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 β}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 β}{2} : β

\frac{2 β}{2}

Divide the numbers:

\frac{2}{2} = 1

= β

Simplify

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 3 0^{\circ} + 18 0^{\circ} n

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}

\frac{6 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 3 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 3 0^{\circ} + 18 0^{\circ} n

β = 3 0^{\circ} + 18 0^{\circ} n

β = 3 0^{\circ} + 18 0^{\circ} n

β = 3 0^{\circ} + 18 0^{\circ} n

Solve

2 β = 30 0^{\circ} + 36 0^{\circ} n : β = 15 0^{\circ} + 18 0^{\circ} n

2 β = 30 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 β = 30 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 β}{2} = \frac{30 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 β}{2} = \frac{30 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 β}{2} : β

\frac{2 β}{2}

Divide the numbers:

\frac{2}{2} = 1

= β

Simplify

\frac{30 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 15 0^{\circ} + 18 0^{\circ} n

\frac{30 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{30 0 ^{\circ}}{2} = 15 0^{\circ}

\frac{30 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{90 0 ^{\circ}}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 15 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 15 0^{\circ} + 18 0^{\circ} n

β = 15 0^{\circ} + 18 0^{\circ} n

β = 15 0^{\circ} + 18 0^{\circ} n

β = 15 0^{\circ} + 18 0^{\circ} n

β = 3 0^{\circ} + 18 0^{\circ} n, β = 15 0^{\circ} + 18 0^{\circ} n

Solutions for the range

- 36 0^{\circ} \leq β \leq 36 0^{\circ}

β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}

cos (2 β) = - 1, - 36 0^{\circ} \leq β \leq 36 0^{\circ} : β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

cos (2 β) = - 1, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

General solutions for

cos (2 β) = - 1

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 β = 18 0^{\circ} + 36 0^{\circ} n

2 β = 18 0^{\circ} + 36 0^{\circ} n

Solve

2 β = 18 0^{\circ} + 36 0^{\circ} n : β = 9 0^{\circ} + 18 0^{\circ} n

2 β = 18 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 β = 18 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 β}{2} = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{2}

Simplify

β = 9 0^{\circ} + 18 0^{\circ} n

β = 9 0^{\circ} + 18 0^{\circ} n

β = 9 0^{\circ} + 18 0^{\circ} n

Solutions for the range

- 36 0^{\circ} \leq β \leq 36 0^{\circ}

β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

Combine all the solutions

β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}, β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 ?
The general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 is