What is the general solution for 6cos(2θ)=18cos(θ)-12 ?

The general solution for 6cos(2θ)=18cos(θ)-12 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6cos(2θ)=18cos(θ)-12

Solution

6 cos (2 θ) = 18 cos (θ) - 12

Solution

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

Degrees

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

Solution steps

6 cos (2 θ) = 18 cos (θ) - 12

Subtract

18 cos (θ) - 12

from both sides

6 cos (2 θ) - 18 cos (θ) + 12 = 0

Rewrite using trig identities

12 - 18 cos (θ) + 6 cos (2 θ)

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= 12 - 18 cos (θ) + 6 (2 cos^{2} (θ) - 1)

Simplify

12 - 18 cos (θ) + 6 (2 cos^{2} (θ) - 1) : 12 cos^{2} (θ) - 18 cos (θ) + 6

12 - 18 cos (θ) + 6 (2 cos^{2} (θ) - 1)

Expand

6 (2 cos^{2} (θ) - 1) : 12 cos^{2} (θ) - 6

6 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = 2 cos^{2} (θ), c = 1

= 6 \cdot 2 cos^{2} (θ) - 6 \cdot 1

Simplify

6 \cdot 2 cos^{2} (θ) - 6 \cdot 1 : 12 cos^{2} (θ) - 6

6 \cdot 2 cos^{2} (θ) - 6 \cdot 1

Multiply the numbers:

6 \cdot 2 = 12

= 12 cos^{2} (θ) - 6 \cdot 1

Multiply the numbers:

6 \cdot 1 = 6

= 12 cos^{2} (θ) - 6

= 12 cos^{2} (θ) - 6

= 12 - 18 cos (θ) + 12 cos^{2} (θ) - 6

Simplify

12 - 18 cos (θ) + 12 cos^{2} (θ) - 6 : 12 cos^{2} (θ) - 18 cos (θ) + 6

12 - 18 cos (θ) + 12 cos^{2} (θ) - 6

Group like terms

= - 18 cos (θ) + 12 cos^{2} (θ) + 12 - 6

Add/Subtract the numbers:

12 - 6 = 6

= 12 cos^{2} (θ) - 18 cos (θ) + 6

= 12 cos^{2} (θ) - 18 cos (θ) + 6

= 12 cos^{2} (θ) - 18 cos (θ) + 6

6 + 12 cos^{2} (θ) - 18 cos (θ) = 0

Solve by substitution

6 + 12 cos^{2} (θ) - 18 cos (θ) = 0

Let:

cos (θ) = u

6 + 12 u^{2} - 18 u = 0

6 + 12 u^{2} - 18 u = 0 : u = 1, u = \frac{1}{2}

6 + 12 u^{2} - 18 u = 0

Write in the standard form

a x^{2} + b x + c = 0

12 u^{2} - 18 u + 6 = 0

Solve with the quadratic formula

12 u^{2} - 18 u + 6 = 0

Quadratic Equation Formula:

For

a = 12, b = - 18, c = 6

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 12 \cdot 6}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 12 \cdot 6}{2 \cdot 12}

(- 18)^{2} - 4 \cdot 12 \cdot 6 = 6

(- 18)^{2} - 4 \cdot 12 \cdot 6

Apply exponent rule:

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

n

is even

(- 18)^{2} = 1 8^{2}

= 1 8^{2} - 4 \cdot 12 \cdot 6

Multiply the numbers:

4 \cdot 12 \cdot 6 = 288

= 1 8^{2} - 288

1 8^{2} = 324

= 324 - 288

Subtract the numbers:

324 - 288 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 \cdot 12}

Separate the solutions

u_{1} = \frac{- ( - 18 ) + 6}{2 \cdot 12}, u_{2} = \frac{- ( - 18 ) - 6}{2 \cdot 12}

u = \frac{- ( - 18 ) + 6}{2 \cdot 12} : 1

\frac{- ( - 18 ) + 6}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{18 + 6}{2 \cdot 12}

Add the numbers:

18 + 6 = 24

= \frac{24}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{24}{24}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 18 ) - 6}{2 \cdot 12} : \frac{1}{2}

\frac{- ( - 18 ) - 6}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{18 - 6}{2 \cdot 12}

Subtract the numbers:

18 - 6 = 12

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

Multiply the numbers:

2 \cdot 12 = 24

= \frac{12}{24}

Cancel the common factor:

12

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 1

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}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{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}

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

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

Combine all the solutions

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

Popular Examples

cos(2/3 x-pi/2)=1

cos (\frac{2}{3} x - \frac{π}{2}) = 1

tan^2(x)sin(x)=sin(x),0<= x<= 2pi

tan^{2} (x) sin (x) = sin (x), 0 \leq x \leq 2 π

-3sin(t)=0

- 3 sin (t) = 0

solvefor x,0=cos(x)-sin(x)solve for

x, 0 = cos (x) - sin (x)

3cos^2(θ)+7cos(θ)-8=0

3 cos^{2} (θ) + 7 cos (θ) - 8 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 6cos(2θ)=18cos(θ)-12 ?
The general solution for 6cos(2θ)=18cos(θ)-12 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin