What is the general solution for 4(1+sin(θ))=cos^2(θ) ?

The general solution for 4(1+sin(θ))=cos^2(θ) is θ=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4(1+sin(θ))=cos^2(θ)

Solution

4 (1 + sin (θ)) = cos^{2} (θ)

Solution

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

4 (1 + sin (θ)) = cos^{2} (θ)

Subtract

cos^{2} (θ)

from both sides

4 (1 + sin (θ)) - cos^{2} (θ) = 0

Rewrite using trig identities

- cos^{2} (θ) + (1 + sin (θ)) \cdot 4

Use the Pythagorean identity:

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

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

= - (1 - sin^{2} (θ)) + (1 + sin (θ)) \cdot 4

Simplify

- (1 - sin^{2} (θ)) + (1 + sin (θ)) \cdot 4 : sin^{2} (θ) + 4 sin (θ) + 3

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

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

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

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + sin^{2} (θ)

= - 1 + sin^{2} (θ) + (1 + sin (θ)) \cdot 4

Expand

4 (1 + sin (θ)) : 4 + 4 sin (θ)

4 (1 + sin (θ))

Apply the distributive law:

a (b + c) = ab + a c

a = 4, b = 1, c = sin (θ)

= 4 \cdot 1 + 4 sin (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 + 4 sin (θ)

= - 1 + sin^{2} (θ) + 4 + 4 sin (θ)

Simplify

- 1 + sin^{2} (θ) + 4 + 4 sin (θ) : sin^{2} (θ) + 4 sin (θ) + 3

- 1 + sin^{2} (θ) + 4 + 4 sin (θ)

Group like terms

= sin^{2} (θ) + 4 sin (θ) - 1 + 4

Add/Subtract the numbers:

- 1 + 4 = 3

= sin^{2} (θ) + 4 sin (θ) + 3

= sin^{2} (θ) + 4 sin (θ) + 3

= sin^{2} (θ) + 4 sin (θ) + 3

3 + sin^{2} (θ) + 4 sin (θ) = 0

Solve by substitution

3 + sin^{2} (θ) + 4 sin (θ) = 0

Let:

sin (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 4, c = 3

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

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

4^{2} - 4 \cdot 1 \cdot 3 = 2

4^{2} - 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

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

Separate the solutions

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

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

\frac{- 4 + 2}{2 \cdot 1}

Add/Subtract the numbers:

- 4 + 2 = - 2

= \frac{- 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 4 - 2}{2 \cdot 1} : - 3

\frac{- 4 - 2}{2 \cdot 1}

Subtract the numbers:

- 4 - 2 = - 6

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6}{2}

Apply the fraction rule:

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

= - \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3

The solutions to the quadratic equation are:

u = - 1, u = - 3

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = - 3

sin (θ) = - 1, sin (θ) = - 3

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

sin (θ) = - 3 :

No Solution

sin (θ) = - 3

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

Graph

Popular Examples

4csc(θ)+4=0

4 csc (θ) + 4 = 0

4cos^2(4x)-3=0

4 cos^{2} (4 x) - 3 = 0

tan^2(x)=cot(x)

tan^{2} (x) = cot (x)

2sin(x)cos(x)sin(x)-cos(x)=0

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

tan^2(θ)=0

tan^{2} (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 4(1+sin(θ))=cos^2(θ) ?
The general solution for 4(1+sin(θ))=cos^2(θ) is θ=(3pi)/2+2pin