What is the general solution for tan^2(x)-1=4cos(x) ?

The general solution for tan^2(x)-1=4cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(x)-1=4cos(x)

Solution

tan^{2} (x) - 1 = 4 cos (x)

Solution

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

tan^{2} (x) - 1 = 4 cos (x)

Square both sides

(tan^{2} (x) - 1)^{2} = (4 cos (x))^{2}

Subtract

(4 cos (x))^{2}

from both sides

(tan^{2} (x) - 1)^{2} - 16 cos^{2} (x) = 0

Rewrite using trig identities

(- 1 + tan^{2} (x))^{2} - 16 cos^{2} (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= (- 1 + (\frac{sin ( x )}{cos ( x )})^{2})^{2} - 16 cos^{2} (x)

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= (- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 cos^{2} (x)

(- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 cos^{2} (x) = 0

Factor

(- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 cos^{2} (x) : (- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x)) (- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x))

(- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 cos^{2} (x)

Rewrite

16 cos^{2} (x)

(2^{2} cos (x))^{2}

16 cos^{2} (x)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

16 = (2^{2})^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Refine

= (\frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x) - 1) (\frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x) - 1)

(- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x)) (- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x)) = 0

Solving each part separately

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x) = 0 or - 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x) = 0

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x) = 0

Simplify

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x) : \frac{- cos ^{2} ( x ) + sin ^{2} ( x ) + 4 cos ^{3} ( x )}{cos ^{2} ( x )}

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x)

Convert element to fraction:

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}, 4 cos (x) = \frac{4 cos ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

= - \frac{1 \cdot cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + \frac{4 cos ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot cos ^{2} ( x ) + sin ^{2} ( x ) + 4 cos ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

- 1 \cdot cos^{2} (x) + sin^{2} (x) + 4 cos (x) cos^{2} (x) = - cos^{2} (x) + sin^{2} (x) + 4 cos^{3} (x)

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

1 \cdot cos^{2} (x) = cos^{2} (x)

1 \cdot cos^{2} (x)

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

= cos^{2} (x)

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

4 cos (x) cos^{2} (x)

Apply exponent rule:

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

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

= 4 cos^{1 + 2} (x)

Add the numbers:

1 + 2 = 3

= 4 cos^{3} (x)

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

= \frac{- cos ^{2} ( x ) + sin ^{2} ( x ) + 4 cos ^{3} ( x )}{cos ^{2} ( x )}

\frac{- cos ^{2} ( x ) + sin ^{2} ( x ) + 4 cos ^{3} ( x )}{cos ^{2} ( x )} = 0

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

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

Rewrite using trig identities

- cos^{2} (x) + sin^{2} (x) + 4 cos^{3} (x)

Use the Pythagorean identity:

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

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

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

Simplify

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

- cos^{2} (x) + 1 - cos^{2} (x) + 4 cos^{3} (x)

Group like terms

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

Add similar elements:

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

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

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

1 - 2 cos^{2} (x) + 4 cos^{3} (x) = 0

Solve by substitution

1 - 2 cos^{2} (x) + 4 cos^{3} (x) = 0

Let:

cos (x) = u

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

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

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

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

Factor

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

4 u^{3} - 2 u^{2} + 1

Use the rational root theorem

a_{0} = 1, a_{n} = 4

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4}

- \frac{1}{2}

is a root of the expression, so factor out

2 u + 1

= (2 u + 1) \frac{4 u ^{3} - 2 u ^{2} + 1}{2 u + 1}

\frac{4 u ^{3} - 2 u ^{2} + 1}{2 u + 1} = 2 u^{2} - 2 u + 1

\frac{4 u ^{3} - 2 u ^{2} + 1}{2 u + 1}

Divide

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

Divide the leading coefficients of the numerator

4 u^{3} - 2 u^{2} + 1

and the divisor

2 u + 1 : \frac{4 u ^{3}}{2 u} = 2 u^{2}

Quotient = 2 u^{2}

Multiply

2 u + 1

2 u^{2} : 4 u^{3} + 2 u^{2}

Subtract

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

from

4 u^{3} - 2 u^{2} + 1

to get new remainder

Remainder = - 4 u^{2} + 1

Therefore

\frac{4 u ^{3} - 2 u ^{2} + 1}{2 u + 1} = 2 u^{2} + \frac{- 4 u ^{2} + 1}{2 u + 1}

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

Divide

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

Divide the leading coefficients of the numerator

- 4 u^{2} + 1

and the divisor

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

Quotient = - 2 u

Multiply

2 u + 1

- 2 u : - 4 u^{2} - 2 u

Subtract

- 4 u^{2} - 2 u

from

- 4 u^{2} + 1

to get new remainder

Remainder = 2 u + 1

Therefore

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

= 2 u^{2} - 2 u + \frac{2 u + 1}{2 u + 1}

Divide

\frac{2 u + 1}{2 u + 1} : \frac{2 u + 1}{2 u + 1} = 1

Divide the leading coefficients of the numerator

2 u + 1

and the divisor

2 u + 1 : \frac{2 u}{2 u} = 1

Quotient = 1

Multiply

2 u + 1

1 : 2 u + 1

Subtract

2 u + 1

from

2 u + 1

to get new remainder

Remainder = 0

Therefore

\frac{2 u + 1}{2 u + 1} = 1

= 2 u^{2} - 2 u + 1

= (2 u + 1) (2 u^{2} - 2 u + 1)

(2 u + 1) (2 u^{2} - 2 u + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 u + 1 = 0 or 2 u^{2} - 2 u + 1 = 0

Solve

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

2 u + 1 = 0

Move

1

to the right side

2 u + 1 = 0

Subtract

1

from both sides

2 u + 1 - 1 = 0 - 1

Simplify

2 u = - 1

2 u = - 1

Divide both sides by

2

2 u = - 1

Divide both sides by

2

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

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Solve

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

2 u^{2} - 2 u + 1 = 0

Solve with the quadratic formula

2 u^{2} - 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 2, c = 1

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

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 1 : 2 i

(- 2)^{2} - 4 \cdot 2 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} - 8

Apply imaginary number rule:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

Add/Subtract the numbers:

- 4 + 8 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 i

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

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 i}{2 \cdot 2}, u_{2} = \frac{- ( - 2 ) - 2 i}{2 \cdot 2}

u = \frac{- ( - 2 ) + 2 i}{2 \cdot 2} : \frac{1}{2} + i \frac{1}{2}

\frac{- ( - 2 ) + 2 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{2 + 2 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 + 2 i}{4}

Factor

2 + 2 i : 2 (1 + i)

2 + 2 i

Rewrite as

= 2 \cdot 1 + 2 i

Factor out common term

2

= 2 (1 + i)

= \frac{2 ( 1 + i )}{4}

Cancel the common factor:

2

= \frac{1 + i}{2}

Rewrite

\frac{1 + i}{2}

in standard complex form:

\frac{1}{2} + \frac{1}{2} i

\frac{1 + i}{2}

Apply the fraction rule:

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

\frac{1 + i}{2} = \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{1}{2} i

u = \frac{- ( - 2 ) - 2 i}{2 \cdot 2} : \frac{1}{2} - i \frac{1}{2}

\frac{- ( - 2 ) - 2 i}{2 \cdot 2}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

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

Factor

2 - 2 i : 2 (1 - i)

2 - 2 i

Rewrite as

= 2 \cdot 1 - 2 i

Factor out common term

2

= 2 (1 - i)

= \frac{2 ( 1 - i )}{4}

Cancel the common factor:

2

= \frac{1 - i}{2}

Rewrite

\frac{1 - i}{2}

in standard complex form:

\frac{1}{2} - \frac{1}{2} i

\frac{1 - i}{2}

Apply the fraction rule:

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

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

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

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

The solutions to the quadratic equation are:

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

The solutions are

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

Substitute back

u = cos (x)

cos (x) = - \frac{1}{2}, cos (x) = \frac{1}{2} + i \frac{1}{2}, cos (x) = \frac{1}{2} - i \frac{1}{2}

cos (x) = - \frac{1}{2}, cos (x) = \frac{1}{2} + i \frac{1}{2}, cos (x) = \frac{1}{2} - i \frac{1}{2}

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

cos (x) = - \frac{1}{2}

General solutions for

cos (x) = - \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}

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

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

cos (x) = \frac{1}{2} + i \frac{1}{2} :

No Solution

cos (x) = \frac{1}{2} + i \frac{1}{2}

No Solution

cos (x) = \frac{1}{2} - i \frac{1}{2} :

No Solution

cos (x) = \frac{1}{2} - i \frac{1}{2}

No Solution

Combine all the solutions

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

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x) = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x) = 0

Simplify

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x) : \frac{- cos ^{2} ( x ) + sin ^{2} ( x ) - 4 cos ^{3} ( x )}{cos ^{2} ( x )}

- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x)

Convert element to fraction:

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}, 4 cos (x) = \frac{4 cos ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

= - \frac{1 \cdot cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - \frac{4 cos ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot cos ^{2} ( x ) + sin ^{2} ( x ) - 4 cos ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

- 1 \cdot cos^{2} (x) + sin^{2} (x) - 4 cos (x) cos^{2} (x) = - cos^{2} (x) + sin^{2} (x) - 4 cos^{3} (x)

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

1 \cdot cos^{2} (x) = cos^{2} (x)

1 \cdot cos^{2} (x)

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

= cos^{2} (x)

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

4 cos (x) cos^{2} (x)

Apply exponent rule:

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

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

= 4 cos^{1 + 2} (x)

Add the numbers:

1 + 2 = 3

= 4 cos^{3} (x)

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

= \frac{- cos ^{2} ( x ) + sin ^{2} ( x ) - 4 cos ^{3} ( x )}{cos ^{2} ( x )}

\frac{- cos ^{2} ( x ) + sin ^{2} ( x ) - 4 cos ^{3} ( x )}{cos ^{2} ( x )} = 0

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

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

Rewrite using trig identities

- cos^{2} (x) + sin^{2} (x) - 4 cos^{3} (x)

Use the Pythagorean identity:

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

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

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

Simplify

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

- cos^{2} (x) + 1 - cos^{2} (x) - 4 cos^{3} (x)

Group like terms

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

Add similar elements:

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

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

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

1 - 2 cos^{2} (x) - 4 cos^{3} (x) = 0

Solve by substitution

1 - 2 cos^{2} (x) - 4 cos^{3} (x) = 0

Let:

cos (x) = u

1 - 2 u^{2} - 4 u^{3} = 0

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

1 - 2 u^{2} - 4 u^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

Factor

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

- 4 u^{3} - 2 u^{2} + 1

Factor out common term

- 1

= - (4 u^{3} + 2 u^{2} - 1)

Factor

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

4 u^{3} + 2 u^{2} - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 4

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4}

\frac{1}{2}

is a root of the expression, so factor out

2 u - 1

= (2 u - 1) \frac{4 u ^{3} + 2 u ^{2} - 1}{2 u - 1}

\frac{4 u ^{3} + 2 u ^{2} - 1}{2 u - 1} = 2 u^{2} + 2 u + 1

\frac{4 u ^{3} + 2 u ^{2} - 1}{2 u - 1}

Divide

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

Divide the leading coefficients of the numerator

4 u^{3} + 2 u^{2} - 1

and the divisor

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

Quotient = 2 u^{2}

Multiply

2 u - 1

2 u^{2} : 4 u^{3} - 2 u^{2}

Subtract

4 u^{3} - 2 u^{2}

from

4 u^{3} + 2 u^{2} - 1

to get new remainder

Remainder = 4 u^{2} - 1

Therefore

\frac{4 u ^{3} + 2 u ^{2} - 1}{2 u - 1} = 2 u^{2} + \frac{4 u ^{2} - 1}{2 u - 1}

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

Divide

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

Divide the leading coefficients of the numerator

4 u^{2} - 1

and the divisor

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

Quotient = 2 u

Multiply

2 u - 1

2 u : 4 u^{2} - 2 u

Subtract

4 u^{2} - 2 u

from

4 u^{2} - 1

to get new remainder

Remainder = 2 u - 1

Therefore

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

= 2 u^{2} + 2 u + \frac{2 u - 1}{2 u - 1}

Divide

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

Divide the leading coefficients of the numerator

2 u - 1

and the divisor

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

Quotient = 1

Multiply

2 u - 1

1 : 2 u - 1

Subtract

2 u - 1

from

2 u - 1

to get new remainder

Remainder = 0

Therefore

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

= 2 u^{2} + 2 u + 1

= 2 u^{2} + 2 u + 1

= (2 u - 1) (2 u^{2} + 2 u + 1)

= - (2 u - 1) (2 u^{2} + 2 u + 1)

- (2 u - 1) (2 u^{2} + 2 u + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 u - 1 = 0 or 2 u^{2} + 2 u + 1 = 0

Solve

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

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

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Solve

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 2, c = 1

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

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

Simplify

2^{2} - 4 \cdot 2 \cdot 1 : 2 i

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} - 8

Apply imaginary number rule:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

Add/Subtract the numbers:

- 4 + 8 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 i

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

Separate the solutions

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

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

\frac{- 2 + 2 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 i}{4}

Factor

- 2 + 2 i : 2 (- 1 + i)

- 2 + 2 i

Rewrite as

= - 2 \cdot 1 + 2 i

Factor out common term

2

= 2 (- 1 + i)

= \frac{2 ( - 1 + i )}{4}

Cancel the common factor:

2

= \frac{- 1 + i}{2}

Rewrite

\frac{- 1 + i}{2}

in standard complex form:

- \frac{1}{2} + \frac{1}{2} i

\frac{- 1 + i}{2}

Apply the fraction rule:

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

\frac{- 1 + i}{2} = - \frac{1}{2} + \frac{i}{2}

= - \frac{1}{2} + \frac{i}{2}

= - \frac{1}{2} + \frac{1}{2} i

u = \frac{- 2 - 2 i}{2 \cdot 2} : - \frac{1}{2} - i \frac{1}{2}

\frac{- 2 - 2 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

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

Factor

- 2 - 2 i : - 2 (1 + i)

- 2 - 2 i

Rewrite as

= - 2 \cdot 1 - 2 i

Factor out common term

2

= - 2 (1 + i)

= - \frac{2 ( 1 + i )}{4}

Cancel the common factor:

2

= - \frac{1 + i}{2}

Rewrite

- \frac{1 + i}{2}

in standard complex form:

- \frac{1}{2} - \frac{1}{2} i

- \frac{1 + i}{2}

Apply the fraction rule:

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

\frac{1 + i}{2} = - (\frac{1}{2}) - (\frac{i}{2})

= - (\frac{1}{2}) - (\frac{i}{2})

Remove parentheses:

(a) = a

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

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

The solutions to the quadratic equation are:

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

The solutions are

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

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2} + i \frac{1}{2}, cos (x) = - \frac{1}{2} - i \frac{1}{2}

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2} + i \frac{1}{2}, cos (x) = - \frac{1}{2} - i \frac{1}{2}

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \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}

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

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

cos (x) = - \frac{1}{2} + i \frac{1}{2} :

No Solution

cos (x) = - \frac{1}{2} + i \frac{1}{2}

No Solution

cos (x) = - \frac{1}{2} - i \frac{1}{2} :

No Solution

cos (x) = - \frac{1}{2} - i \frac{1}{2}

No Solution

Combine all the solutions

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

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan^{2} (x) - 1 = 4 cos (x)

Remove the ones that don't agree with the equation.

Check the solution

\frac{2 π}{3} + 2 πn :

False

\frac{2 π}{3} + 2 πn

Plug in

n = 1

\frac{2 π}{3} + 2 π 1

For

tan^{2} (x) - 1 = 4 cos (x)

plug in

x = \frac{2 π}{3} + 2 π 1

tan^{2} (\frac{2 π}{3} + 2 π 1) - 1 = 4 cos (\frac{2 π}{3} + 2 π 1)

Refine

2 = - 2

\Rightarrow False

Check the solution

\frac{4 π}{3} + 2 πn :

False

\frac{4 π}{3} + 2 πn

Plug in

n = 1

\frac{4 π}{3} + 2 π 1

For

tan^{2} (x) - 1 = 4 cos (x)

plug in

x = \frac{4 π}{3} + 2 π 1

tan^{2} (\frac{4 π}{3} + 2 π 1) - 1 = 4 cos (\frac{4 π}{3} + 2 π 1)

Refine

2 = - 2

\Rightarrow False

Check the solution

\frac{π}{3} + 2 πn :

True

\frac{π}{3} + 2 πn

Plug in

n = 1

\frac{π}{3} + 2 π 1

For

tan^{2} (x) - 1 = 4 cos (x)

plug in

x = \frac{π}{3} + 2 π 1

tan^{2} (\frac{π}{3} + 2 π 1) - 1 = 4 cos (\frac{π}{3} + 2 π 1)

Refine

2 = 2

\Rightarrow True

Check the solution

\frac{5 π}{3} + 2 πn :

True

\frac{5 π}{3} + 2 πn

Plug in

n = 1

\frac{5 π}{3} + 2 π 1

For

tan^{2} (x) - 1 = 4 cos (x)

plug in

x = \frac{5 π}{3} + 2 π 1

tan^{2} (\frac{5 π}{3} + 2 π 1) - 1 = 4 cos (\frac{5 π}{3} + 2 π 1)

Refine

2 = 2

\Rightarrow True

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

Popular Examples

(10)/(sin(30))= 3/(sin(b))

\frac{10}{sin ( 3 0 ^{\circ} )} = \frac{3}{sin ( b )}

tan(θ/2-pi/6)=-1,0<= θ<2pi

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

arctan(x+2)-arctan(x+1)= pi/4

arctan (x + 2) - arctan (x + 1) = \frac{π}{4}

3cos(2x)=1

3 cos (2 x) = 1

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

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

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)-1=4cos(x) ?
The general solution for tan^2(x)-1=4cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin