What is the general solution for -2+cos(θ)=-2+6/5*cos(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-2+cos(θ)=-2+6/5*cos(θ)

Solution

- 2 + cos (θ) = - 2 + \frac{6}{5} \cdot cos (θ)

Solution

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

- 2 + cos (θ) = - 2 + \frac{6}{5} cos (θ)

Solve by substitution

- 2 + cos (θ) = - 2 + \frac{6}{5} cos (θ)

Let:

cos (θ) = u

- 2 + u = - 2 + \frac{6}{5} u

- 2 + u = - 2 + \frac{6}{5} u : u = 0

- 2 + u = - 2 + \frac{6}{5} u

Add

2

to both sides

- 2 + u + 2 = - 2 + \frac{6}{5} u + 2

Simplify

u = \frac{6}{5} u

Multiply both sides by

5

u = \frac{6}{5} u

Multiply both sides by

5

u \cdot 5 = \frac{6}{5} u \cdot 5

Simplify

u \cdot 5 = \frac{6}{5} u \cdot 5

Simplify

u \cdot 5 : 5 u

u \cdot 5

Apply the commutative law:

u \cdot 5 = 5 u

5 u

Simplify

\frac{6}{5} u \cdot 5 : 6 u

\frac{6}{5} u \cdot 5

Multiply fractions:

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

= \frac{6 \cdot 5}{5} u

Cancel the common factor:

5

= u \cdot 6

5 u = 6 u

5 u = 6 u

5 u = 6 u

Move

6 u

to the left side

5 u = 6 u

Subtract

6 u

from both sides

5 u - 6 u = 6 u - 6 u

Simplify

- u = 0

- u = 0

Divide both sides by

- 1

- u = 0

Divide both sides by

- 1

\frac{- u}{- 1} = \frac{0}{- 1}

Simplify

u = 0

u = 0

Substitute back

u = cos (θ)

cos (θ) = 0

cos (θ) = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

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

Combine all the solutions

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

Popular Examples

tan^2(y)+3-2sec^2(y)=0

tan^{2} (y) + 3 - 2 sec^{2} (y) = 0

500=538.5cos(x-10)

500 = 538.5 cos (x - 1 0^{\circ})

-2cos^2(x)=3(sin(x)-1)

- 2 cos^{2} (x) = 3 (sin (x) - 1)

cos(x)tan(x)=sin(2x)

cos (x) tan (x) = sin (2 x)

2|sin(4θ)|=0

2 ∣ sin (4 θ) ∣ = 0

Frequently Asked Questions (FAQ)

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