What is the general solution for 2sqrt(3)cos(2θ)=-3 ?

The general solution for 2sqrt(3)cos(2θ)=-3 is θ=(5pi)/(12)+pin,θ=(7pi)/(12)+pin

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2sqrt(3)cos(2θ)=-3

Solution

23 cos (2 θ) = - 3

Solution

θ = \frac{5 π}{12} + πn, θ = \frac{7 π}{12} + πn

Degrees

θ = 7 5^{\circ} + 18 0^{\circ} n, θ = 10 5^{\circ} + 18 0^{\circ} n

Solution steps

23 cos (2 θ) = - 3

Divide both sides by

23

23 cos (2 θ) = - 3

Divide both sides by

23

\frac{2 3 cos ( 2 θ )}{2 3} = \frac{- 3}{2 3}

Simplify

\frac{2 3 cos ( 2 θ )}{2 3} = \frac{- 3}{2 3}

Simplify

\frac{2 3 cos ( 2 θ )}{2 3} : cos (2 θ)

\frac{2 3 cos ( 2 θ )}{2 3}

Divide the numbers:

\frac{2}{2} = 1

= \frac{3 cos ( 2 θ )}{3}

Cancel the common factor:

3

= cos (2 θ)

Simplify

\frac{- 3}{2 3} : - \frac{3}{2}

\frac{- 3}{2 3}

Apply the fraction rule:

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

= - \frac{3}{2 3}

Apply radical rule:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

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

Apply exponent rule:

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

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

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

Subtract the numbers:

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

= \frac{3 ^{\frac{1}{2}}}{2}

Apply radical rule:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= - \frac{3}{2}

cos (2 θ) = - \frac{3}{2}

cos (2 θ) = - \frac{3}{2}

cos (2 θ) = - \frac{3}{2}

General solutions for

cos (2 θ) = - \frac{3}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 θ = \frac{5 π}{6} + 2 πn, 2 θ = \frac{7 π}{6} + 2 πn

2 θ = \frac{5 π}{6} + 2 πn, 2 θ = \frac{7 π}{6} + 2 πn

Solve

2 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{12} + πn

2 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{5 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{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + 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

θ = \frac{5 π}{12} + πn, θ = \frac{7 π}{12} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sqrt(3)cos(2θ)=-3 ?
The general solution for 2sqrt(3)cos(2θ)=-3 is θ=(5pi)/(12)+pin,θ=(7pi)/(12)+pin