What is the general solution for (3+sin^2(θ))/(cos(θ)-2)=3cos(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(3+sin^2(θ))/(cos(θ)-2)=3cos(θ)

Solution

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} = 3 cos (θ)

Solution

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

Degrees

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

Solution steps

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} = 3 cos (θ)

Subtract

3 cos (θ)

from both sides

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} - 3 cos (θ) = 0

Simplify

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} - 3 cos (θ) : \frac{3 + sin ^{2} ( θ ) - 3 cos ( θ ) ( cos ( θ ) - 2 )}{cos ( θ ) - 2}

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} - 3 cos (θ)

Convert element to fraction:

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

= \frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} - \frac{3 cos ( θ ) ( cos ( θ ) - 2 )}{cos ( θ ) - 2}

Since the denominators are equal, combine the fractions:

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

= \frac{3 + sin ^{2} ( θ ) - 3 cos ( θ ) ( cos ( θ ) - 2 )}{cos ( θ ) - 2}

\frac{3 + sin ^{2} ( θ ) - 3 cos ( θ ) ( cos ( θ ) - 2 )}{cos ( θ ) - 2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 + sin^{2} (θ) - 3 cos (θ) (cos (θ) - 2) = 0

Rewrite using trig identities

3 + sin^{2} (θ) - (- 2 + cos (θ)) \cdot 3 cos (θ)

Use the Pythagorean identity:

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

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

= 3 + 1 - cos^{2} (θ) - (- 2 + cos (θ)) \cdot 3 cos (θ)

Simplify

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

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

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

Expand

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

- 3 cos (θ) (- 2 + cos (θ))

Apply the distributive law:

a (b + c) = ab + a c

a = - 3 cos (θ), b = - 2, c = cos (θ)

= - 3 cos (θ) (- 2) + (- 3 cos (θ)) cos (θ)

Apply minus-plus rules

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

= 3 \cdot 2 cos (θ) - 3 cos (θ) cos (θ)

Simplify

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

3 \cdot 2 cos (θ) - 3 cos (θ) cos (θ)

3 \cdot 2 cos (θ) = 6 cos (θ)

3 \cdot 2 cos (θ)

Multiply the numbers:

3 \cdot 2 = 6

= 6 cos (θ)

3 cos (θ) cos (θ) = 3 cos^{2} (θ)

3 cos (θ) cos (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 3 cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 3 cos^{2} (θ)

= 6 cos (θ) - 3 cos^{2} (θ)

= 6 cos (θ) - 3 cos^{2} (θ)

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

= - 4 cos^{2} (θ) + 6 cos (θ) + 3 + 1

Add the numbers:

3 + 1 = 4

= 6 cos (θ) - 4 cos^{2} (θ) + 4

= 6 cos (θ) - 4 cos^{2} (θ) + 4

= 6 cos (θ) - 4 cos^{2} (θ) + 4

4 - 4 cos^{2} (θ) + 6 cos (θ) = 0

Solve by substitution

4 - 4 cos^{2} (θ) + 6 cos (θ) = 0

Let:

cos (θ) = u

4 - 4 u^{2} + 6 u = 0

4 - 4 u^{2} + 6 u = 0 : u = - \frac{1}{2}, u = 2

4 - 4 u^{2} + 6 u = 0

Write in the standard form

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

- 4 u^{2} + 6 u + 4 = 0

Solve with the quadratic formula

- 4 u^{2} + 6 u + 4 = 0

Quadratic Equation Formula:

For

a = - 4, b = 6, c = 4

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

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

6^{2} - 4 (- 4) \cdot 4 = 10

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

Apply rule

- (- a) = a

= 6^{2} + 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 6^{2} + 64

6^{2} = 36

= 36 + 64

Add the numbers:

36 + 64 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 6 \pm 10}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 6 + 10}{2 ( - 4 )}, u_{2} = \frac{- 6 - 10}{2 ( - 4 )}

u = \frac{- 6 + 10}{2 ( - 4 )} : - \frac{1}{2}

\frac{- 6 + 10}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 10}{- 2 \cdot 4}

Add/Subtract the numbers:

- 6 + 10 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{- 8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

u = \frac{- 6 - 10}{2 ( - 4 )} : 2

\frac{- 6 - 10}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 10}{- 2 \cdot 4}

Subtract the numbers:

- 6 - 10 = - 16

= \frac{- 16}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 16}{- 8}

Apply the fraction rule:

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

= \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= 2

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{2}, cos (θ) = 2

cos (θ) = - \frac{1}{2}, cos (θ) = 2

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

cos (θ) = 2 :

No Solution

cos (θ) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Popular Examples

cos(a)=0

cos (a) = 0

cos(3x)=(-sqrt(3))/2

cos (3 x) = \frac{- 3}{2}

2cos(2θ)-10sin(θ)-2=-6sin(θ)-3

2 cos (2 θ) - 10 sin (θ) - 2 = - 6 sin (θ) - 3

cos(θ)=csc(θ)

cos (θ) = csc (θ)

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

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

Frequently Asked Questions (FAQ)

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