What is the general solution for 2+2sin(θ)=4cos^2(θ) ?

The general solution for 2+2sin(θ)=4cos^2(θ) is θ= pi/6+2pin,θ=(5pi)/6+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+2sin(θ)=4cos^2(θ)

Solution

2 + 2 sin (θ) = 4 cos^{2} (θ)

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 + 2 sin (θ) = 4 cos^{2} (θ)

Subtract

4 cos^{2} (θ)

from both sides

2 + 2 sin (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

2 + 2 sin (θ) - 4 cos^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= 2 + 2 sin (θ) - 4 (1 - sin^{2} (θ))

Simplify

2 + 2 sin (θ) - 4 (1 - sin^{2} (θ)) : 4 sin^{2} (θ) + 2 sin (θ) - 2

2 + 2 sin (θ) - 4 (1 - sin^{2} (θ))

Expand

- 4 (1 - sin^{2} (θ)) : - 4 + 4 sin^{2} (θ)

- 4 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (θ)

= - 4 \cdot 1 - (- 4) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (θ)

= 2 + 2 sin (θ) - 4 + 4 sin^{2} (θ)

Simplify

2 + 2 sin (θ) - 4 + 4 sin^{2} (θ) : 4 sin^{2} (θ) + 2 sin (θ) - 2

2 + 2 sin (θ) - 4 + 4 sin^{2} (θ)

Group like terms

= 2 sin (θ) + 4 sin^{2} (θ) + 2 - 4

Add/Subtract the numbers:

2 - 4 = - 2

= 4 sin^{2} (θ) + 2 sin (θ) - 2

= 4 sin^{2} (θ) + 2 sin (θ) - 2

= 4 sin^{2} (θ) + 2 sin (θ) - 2

- 2 + 2 sin (θ) + 4 sin^{2} (θ) = 0

Solve by substitution

- 2 + 2 sin (θ) + 4 sin^{2} (θ) = 0

Let:

sin (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = 2, c = - 2

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

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

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

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

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

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

Separate the solutions

u_{1} = \frac{- 2 + 6}{2 \cdot 4}, u_{2} = \frac{- 2 - 6}{2 \cdot 4}

u = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

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

Add/Subtract the numbers:

- 2 + 6 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{- 2 - 6}{2 \cdot 4} : - 1

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

Subtract the numbers:

- 2 - 6 = - 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - 1

sin (θ) = \frac{1}{2}, sin (θ) = - 1

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

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

sin (θ) = - 1

General solutions for

sin (θ) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Popular Examples

9tan(x)sin(x)-5sin(x)=0

9 tan (x) sin (x) - 5 sin (x) = 0

solvefor x,y=arctan(x)solve for

x, y = arctan (x)

7cos(x)+7sin(x)tan(x)=14

7 cos (x) + 7 sin (x) tan (x) = 14

cot(θ)+5csc(θ)=6

cot (θ) + 5 csc (θ) = 6

sqrt(3)csc^2(x)+2csc(x)=0

3 csc^{2} (x) + 2 csc (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2+2sin(θ)=4cos^2(θ) ?
The general solution for 2+2sin(θ)=4cos^2(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin