What is the general solution for 14+7cos(2x)=21cos(x) ?

The general solution for 14+7cos(2x)=21cos(x) is x=2pin,x= pi/3+2pin,x=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

14+7cos(2x)=21cos(x)

Solution

14 + 7 cos (2 x) = 21 cos (x)

Solution

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

Degrees

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

Solution steps

14 + 7 cos (2 x) = 21 cos (x)

Subtract

21 cos (x)

from both sides

14 + 7 cos (2 x) - 21 cos (x) = 0

Rewrite using trig identities

14 - 21 cos (x) + 7 cos (2 x)

Use the Double Angle identity:

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

= 14 - 21 cos (x) + 7 (2 cos^{2} (x) - 1)

Simplify

14 - 21 cos (x) + 7 (2 cos^{2} (x) - 1) : 14 cos^{2} (x) - 21 cos (x) + 7

14 - 21 cos (x) + 7 (2 cos^{2} (x) - 1)

Expand

7 (2 cos^{2} (x) - 1) : 14 cos^{2} (x) - 7

7 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 7, b = 2 cos^{2} (x), c = 1

= 7 \cdot 2 cos^{2} (x) - 7 \cdot 1

Simplify

7 \cdot 2 cos^{2} (x) - 7 \cdot 1 : 14 cos^{2} (x) - 7

7 \cdot 2 cos^{2} (x) - 7 \cdot 1

Multiply the numbers:

7 \cdot 2 = 14

= 14 cos^{2} (x) - 7 \cdot 1

Multiply the numbers:

7 \cdot 1 = 7

= 14 cos^{2} (x) - 7

= 14 cos^{2} (x) - 7

= 14 - 21 cos (x) + 14 cos^{2} (x) - 7

Simplify

14 - 21 cos (x) + 14 cos^{2} (x) - 7 : 14 cos^{2} (x) - 21 cos (x) + 7

14 - 21 cos (x) + 14 cos^{2} (x) - 7

Group like terms

= - 21 cos (x) + 14 cos^{2} (x) + 14 - 7

Add/Subtract the numbers:

14 - 7 = 7

= 14 cos^{2} (x) - 21 cos (x) + 7

= 14 cos^{2} (x) - 21 cos (x) + 7

= 14 cos^{2} (x) - 21 cos (x) + 7

7 + 14 cos^{2} (x) - 21 cos (x) = 0

Solve by substitution

7 + 14 cos^{2} (x) - 21 cos (x) = 0

Let:

cos (x) = u

7 + 14 u^{2} - 21 u = 0

7 + 14 u^{2} - 21 u = 0 : u = 1, u = \frac{1}{2}

7 + 14 u^{2} - 21 u = 0

Write in the standard form

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

14 u^{2} - 21 u + 7 = 0

Solve with the quadratic formula

14 u^{2} - 21 u + 7 = 0

Quadratic Equation Formula:

For

a = 14, b = - 21, c = 7

u_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 14 \cdot 7}{2 \cdot 14}

u_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 14 \cdot 7}{2 \cdot 14}

(- 21)^{2} - 4 \cdot 14 \cdot 7 = 7

(- 21)^{2} - 4 \cdot 14 \cdot 7

Apply exponent rule:

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

n

is even

(- 21)^{2} = 2 1^{2}

= 2 1^{2} - 4 \cdot 14 \cdot 7

Multiply the numbers:

4 \cdot 14 \cdot 7 = 392

= 2 1^{2} - 392

2 1^{2} = 441

= 441 - 392

Subtract the numbers:

441 - 392 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 21 ) \pm 7}{2 \cdot 14}

Separate the solutions

u_{1} = \frac{- ( - 21 ) + 7}{2 \cdot 14}, u_{2} = \frac{- ( - 21 ) - 7}{2 \cdot 14}

u = \frac{- ( - 21 ) + 7}{2 \cdot 14} : 1

\frac{- ( - 21 ) + 7}{2 \cdot 14}

Apply rule

- (- a) = a

= \frac{21 + 7}{2 \cdot 14}

Add the numbers:

21 + 7 = 28

= \frac{28}{2 \cdot 14}

Multiply the numbers:

2 \cdot 14 = 28

= \frac{28}{28}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 21 ) - 7}{2 \cdot 14} : \frac{1}{2}

\frac{- ( - 21 ) - 7}{2 \cdot 14}

Apply rule

- (- a) = a

= \frac{21 - 7}{2 \cdot 14}

Subtract the numbers:

21 - 7 = 14

= \frac{14}{2 \cdot 14}

Multiply the numbers:

2 \cdot 14 = 28

= \frac{14}{28}

Cancel the common factor:

14

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = \frac{1}{2}

cos (x) = 1, cos (x) = \frac{1}{2}

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

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

General solutions for

cos (x) = \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}

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

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

Combine all the solutions

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

Popular Examples

-8sin^2(θ)-8sin(θ)+12=-5sin(θ)+7

- 8 sin^{2} (θ) - 8 sin (θ) + 12 = - 5 sin (θ) + 7

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

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

sin(3x)=0,0<= x<2pi

sin (3 x) = 0, 0 \leq x < 2 π

2cos^2(x)=sqrt(3)cos(x)

2 cos^{2} (x) = 3 cos (x)

sin(θ)=-1/5 ,pi<θ<(3pi)/2 ,cos(θ)

sin (θ) = - \frac{1}{5}, π < θ < \frac{3 π}{2}, cos (θ)

Frequently Asked Questions (FAQ)

What is the general solution for 14+7cos(2x)=21cos(x) ?
The general solution for 14+7cos(2x)=21cos(x) is x=2pin,x= pi/3+2pin,x=(5pi)/3+2pin