What is the general solution for 2cos(2θ)-3cos(θ)+12=5cos(θ)+7 ?

The general solution for 2cos(2θ)-3cos(θ)+12=5cos(θ)+7 is 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 >

2cos(2θ)-3cos(θ)+12=5cos(θ)+7

Solution

2 cos (2 θ) - 3 cos (θ) + 12 = 5 cos (θ) + 7

Solution

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

Degrees

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

Solution steps

2 cos (2 θ) - 3 cos (θ) + 12 = 5 cos (θ) + 7

Subtract

5 cos (θ) + 7

from both sides

2 cos (2 θ) - 8 cos (θ) + 5 = 0

Rewrite using trig identities

5 + 2 cos (2 θ) - 8 cos (θ)

Use the Double Angle identity:

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

= 5 + 2 (2 cos^{2} (θ) - 1) - 8 cos (θ)

Simplify

5 + 2 (2 cos^{2} (θ) - 1) - 8 cos (θ) : 4 cos^{2} (θ) - 8 cos (θ) + 3

5 + 2 (2 cos^{2} (θ) - 1) - 8 cos (θ)

Expand

2 (2 cos^{2} (θ) - 1) : 4 cos^{2} (θ) - 2

2 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 2 cos^{2} (θ), c = 1

= 2 \cdot 2 cos^{2} (θ) - 2 \cdot 1

Simplify

2 \cdot 2 cos^{2} (θ) - 2 \cdot 1 : 4 cos^{2} (θ) - 2

2 \cdot 2 cos^{2} (θ) - 2 \cdot 1

Multiply the numbers:

2 \cdot 2 = 4

= 4 cos^{2} (θ) - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 4 cos^{2} (θ) - 2

= 4 cos^{2} (θ) - 2

= 5 + 4 cos^{2} (θ) - 2 - 8 cos (θ)

Simplify

5 + 4 cos^{2} (θ) - 2 - 8 cos (θ) : 4 cos^{2} (θ) - 8 cos (θ) + 3

5 + 4 cos^{2} (θ) - 2 - 8 cos (θ)

Group like terms

= 4 cos^{2} (θ) - 8 cos (θ) + 5 - 2

Add/Subtract the numbers:

5 - 2 = 3

= 4 cos^{2} (θ) - 8 cos (θ) + 3

= 4 cos^{2} (θ) - 8 cos (θ) + 3

= 4 cos^{2} (θ) - 8 cos (θ) + 3

3 + 4 cos^{2} (θ) - 8 cos (θ) = 0

Solve by substitution

3 + 4 cos^{2} (θ) - 8 cos (θ) = 0

Let:

cos (θ) = u

3 + 4 u^{2} - 8 u = 0

3 + 4 u^{2} - 8 u = 0 : u = \frac{3}{2}, u = \frac{1}{2}

3 + 4 u^{2} - 8 u = 0

Write in the standard form

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

4 u^{2} - 8 u + 3 = 0

Solve with the quadratic formula

4 u^{2} - 8 u + 3 = 0

Quadratic Equation Formula:

For

a = 4, b = - 8, c = 3

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

(- 8)^{2} - 4 \cdot 4 \cdot 3 = 4

(- 8)^{2} - 4 \cdot 4 \cdot 3

Apply exponent rule:

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

n

is even

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

= 8^{2} - 4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

Subtract the numbers:

64 - 48 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

u_{1, 2} = \frac{- ( - 8 ) \pm 4}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 4}{2 \cdot 4}, u_{2} = \frac{- ( - 8 ) - 4}{2 \cdot 4}

u = \frac{- ( - 8 ) + 4}{2 \cdot 4} : \frac{3}{2}

\frac{- ( - 8 ) + 4}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{8 + 4}{2 \cdot 4}

Add the numbers:

8 + 4 = 12

= \frac{12}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{12}{8}

Cancel the common factor:

4

= \frac{3}{2}

u = \frac{- ( - 8 ) - 4}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 8 ) - 4}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{8 - 4}{2 \cdot 4}

Subtract the numbers:

8 - 4 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{3}{2}, u = \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{3}{2}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{3}{2}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{3}{2} :

No Solution

cos (θ) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

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

Popular Examples

sin(4x)-sin(x)=0

sin (4 x) - sin (x) = 0

tan(θ)= 7/5

tan (θ) = \frac{7}{5}

sin(t)= 2/5

sin (t) = \frac{2}{5}

tan(x)-1=sec(x)

tan (x) - 1 = sec (x)

0=sec(x)

0 = sec (x)

Frequently Asked Questions (FAQ)

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