English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3cos(2θ)+6=-5cos(θ)+2

Solution

3 cos (2 θ) + 6 = - 5 cos (θ) + 2

Solution

θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Degrees

θ = 109.47122 \dots^{\circ} + 36 0^{\circ} n, θ = - 109.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (2 θ) + 6 = - 5 cos (θ) + 2

Subtract

- 5 cos (θ) + 2

from both sides

3 cos (2 θ) + 5 cos (θ) + 4 = 0

Rewrite using trig identities

4 + 3 cos (2 θ) + 5 cos (θ)

Use the Double Angle identity:

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

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

Simplify

4 + 3 (2 cos^{2} (θ) - 1) + 5 cos (θ) : 6 cos^{2} (θ) + 5 cos (θ) + 1

4 + 3 (2 cos^{2} (θ) - 1) + 5 cos (θ)

Expand

3 (2 cos^{2} (θ) - 1) : 6 cos^{2} (θ) - 3

3 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 2 cos^{2} (θ), c = 1

= 3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

Simplify

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1 : 6 cos^{2} (θ) - 3

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

Multiply the numbers:

3 \cdot 2 = 6

= 6 cos^{2} (θ) - 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 6 cos^{2} (θ) - 3

= 6 cos^{2} (θ) - 3

= 4 + 6 cos^{2} (θ) - 3 + 5 cos (θ)

Simplify

4 + 6 cos^{2} (θ) - 3 + 5 cos (θ) : 6 cos^{2} (θ) + 5 cos (θ) + 1

4 + 6 cos^{2} (θ) - 3 + 5 cos (θ)

Group like terms

= 6 cos^{2} (θ) + 5 cos (θ) + 4 - 3

Add/Subtract the numbers:

4 - 3 = 1

= 6 cos^{2} (θ) + 5 cos (θ) + 1

= 6 cos^{2} (θ) + 5 cos (θ) + 1

= 6 cos^{2} (θ) + 5 cos (θ) + 1

1 + 5 cos (θ) + 6 cos^{2} (θ) = 0

Solve by substitution

1 + 5 cos (θ) + 6 cos^{2} (θ) = 0

Let:

cos (θ) = u

1 + 5 u + 6 u^{2} = 0

1 + 5 u + 6 u^{2} = 0 : u = - \frac{1}{3}, u = - \frac{1}{2}

1 + 5 u + 6 u^{2} = 0

Write in the standard form

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

6 u^{2} + 5 u + 1 = 0

Solve with the quadratic formula

6 u^{2} + 5 u + 1 = 0

Quadratic Equation Formula:

For

a = 6, b = 5, c = 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

5^{2} - 4 \cdot 6 \cdot 1 = 1

5^{2} - 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 5 + 1}{2 \cdot 6}, u_{2} = \frac{- 5 - 1}{2 \cdot 6}

u = \frac{- 5 + 1}{2 \cdot 6} : - \frac{1}{3}

\frac{- 5 + 1}{2 \cdot 6}

Add/Subtract the numbers:

- 5 + 1 = - 4

= \frac{- 4}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 4}{12}

Apply the fraction rule:

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

= - \frac{4}{12}

Cancel the common factor:

4

= - \frac{1}{3}

u = \frac{- 5 - 1}{2 \cdot 6} : - \frac{1}{2}

\frac{- 5 - 1}{2 \cdot 6}

Subtract the numbers:

- 5 - 1 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

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

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

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

cos (θ) = - \frac{1}{3}

Apply trig inverse properties

cos (θ) = - \frac{1}{3}

General solutions for

cos (θ) = - \frac{1}{3}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

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

Combine all the solutions

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Show solutions in decimal form

θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Popular Examples

sin(3x)cos(x)-sin(x)cos(3x)=0

sin (3 x) cos (x) - sin (x) cos (3 x) = 0

sin(x)= 8/12

sin (x) = \frac{8}{12}

2cos(θ)sin(θ)+sqrt(3)cos(θ)=0

2 cos (θ) sin (θ) + 3 cos (θ) = 0

2-2sin(θ)=4

2 - 2 sin (θ) = 4

(sin(x))^2=0

(sin (x))^{2} = 0