What is the general solution for 2cos^2(a)=1+cos(120) ?

The general solution for 2cos^2(a)=1+cos(120) is a=60+360n,a=300+360n,a=120+360n,a=240+360n

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Popular Trigonometry >

2cos^2(a)=1+cos(120)

Solution

2 cos^{2} (a) = 1 + cos (12 0^{\circ})

Solution

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n, a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

Radians

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn, a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn

Solution steps

2 cos^{2} (a) = 1 + cos (12 0^{\circ})

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

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}

= - \frac{1}{2}

= - \frac{1}{2}

2 cos^{2} (a) = 1 + - \frac{1}{2}

Solve by substitution

2 cos^{2} (a) = 1 + - \frac{1}{2}

Let:

cos (a) = u

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

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 u ^{2}}{2} : u^{2}

\frac{2 u ^{2}}{2}

Divide the numbers:

\frac{2}{2} = 1

= u^{2}

Simplify

\frac{1}{2} - \frac{\frac{1}{2}}{2} : \frac{1}{4}

\frac{1}{2} - \frac{\frac{1}{2}}{2}

Apply rule

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

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

Join

1 - \frac{1}{2} : \frac{1}{2}

1 - \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= \frac{1}{2}

= \frac{\frac{1}{2}}{2}

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{4}, u = - \frac{1}{4}

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

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

Simplify

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

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

Substitute back

u = cos (a)

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

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

cos (a) = \frac{1}{2} : a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

cos (a) = \frac{1}{2}

General solutions for

cos (a) = \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}

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

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

cos (a) = - \frac{1}{2} : a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

cos (a) = - \frac{1}{2}

General solutions for

cos (a) = - \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}

a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

Combine all the solutions

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n, a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(a)=1+cos(120) ?
The general solution for 2cos^2(a)=1+cos(120) is a=60+360n,a=300+360n,a=120+360n,a=240+360n