What is the general solution for 3/(tan(x))=(2tan(x))(2cos(x)) ?

The general solution for 3/(tan(x))=(2tan(x))(2cos(x)) is x=0.80515…+2pin,x=2pi-0.80515…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

3/(tan(x))=(2tan(x))(2cos(x))

Solution

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

Solution

x = 0.80515 \dots + 2 πn, x = 2 π - 0.80515 \dots + 2 πn

Degrees

x = 46.13190 \dots^{\circ} + 36 0^{\circ} n, x = 313.86809 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2 tan (x) 2 cos (x)

from both sides

\frac{3}{tan ( x )} - 4 tan (x) cos (x) = 0

Simplify

\frac{3}{tan ( x )} - 4 tan (x) cos (x) : \frac{3 - 4 tan ^{2} ( x ) cos ( x )}{tan ( x )}

\frac{3}{tan ( x )} - 4 tan (x) cos (x)

Convert element to fraction:

4 tan (x) cos (x) = \frac{4 tan ( x ) cos ( x ) tan ( x )}{tan ( x )}

= \frac{3}{tan ( x )} - \frac{4 tan ( x ) cos ( x ) tan ( x )}{tan ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{3 - 4 tan ( x ) cos ( x ) tan ( x )}{tan ( x )}

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

3 - 4 tan (x) cos (x) tan (x)

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

4 tan (x) cos (x) tan (x)

Apply exponent rule:

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

tan (x) tan (x) = tan^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

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

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

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

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

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

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

Express with sin, cos

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

Simplify

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

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

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

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

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

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Add

4 sin^{2} (x)

to both sides

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

Square both sides

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

Subtract

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

from both sides

9 cos^{2} (x) - 16 sin^{4} (x) = 0

Factor

9 cos^{2} (x) - 16 sin^{4} (x) : (3 cos (x) + 4 sin^{2} (x)) (3 cos (x) - 4 sin^{2} (x))

9 cos^{2} (x) - 16 sin^{4} (x)

Rewrite

9 cos^{2} (x) - 16 sin^{4} (x)

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

9 cos^{2} (x) - 16 sin^{4} (x)

Rewrite

9

3^{2}

= 3^{2} cos^{2} (x) - 16 sin^{4} (x)

Rewrite

16

4^{2}

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

Apply exponent rule:

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

sin^{4} (x) = (sin^{2} (x))^{2}

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

4^{2} (sin^{2} (x))^{2} = (4 sin^{2} (x))^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

Solving each part separately

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

3 cos (x) + 4 sin^{2} (x) = 0 : x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

(1 - u^{2}) \cdot 4 + 3 u = 0

(1 - u^{2}) \cdot 4 + 3 u = 0 : u = - \frac{- 3 + 73}{8}, u = \frac{3 + 73}{8}

(1 - u^{2}) \cdot 4 + 3 u = 0

Expand

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

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

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

Expand

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

4 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = u^{2}

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

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = 3, c = 4

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

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

3^{2} - 4 (- 4) \cdot 4 = 73

3^{2} - 4 (- 4) \cdot 4

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 3^{2} + 64

3^{2} = 9

= 9 + 64

Add the numbers:

9 + 64 = 73

= 73

u_{1, 2} = \frac{- 3 \pm 73}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 3 + 73}{2 ( - 4 )}, u_{2} = \frac{- 3 - 73}{2 ( - 4 )}

u = \frac{- 3 + 73}{2 ( - 4 )} : - \frac{- 3 + 73}{8}

\frac{- 3 + 73}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 73}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 3 + 73}{- 8}

Apply the fraction rule:

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

= - \frac{- 3 + 73}{8}

u = \frac{- 3 - 73}{2 ( - 4 )} : \frac{3 + 73}{8}

\frac{- 3 - 73}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 73}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 3 - 73}{- 8}

Apply the fraction rule:

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

- 3 - 73 = - (3 + 73)

= \frac{3 + 73}{8}

The solutions to the quadratic equation are:

u = - \frac{- 3 + 73}{8}, u = \frac{3 + 73}{8}

Substitute back

u = cos (x)

cos (x) = - \frac{- 3 + 73}{8}, cos (x) = \frac{3 + 73}{8}

cos (x) = - \frac{- 3 + 73}{8}, cos (x) = \frac{3 + 73}{8}

cos (x) = - \frac{- 3 + 73}{8} : x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

cos (x) = - \frac{- 3 + 73}{8}

Apply trig inverse properties

cos (x) = - \frac{- 3 + 73}{8}

General solutions for

cos (x) = - \frac{- 3 + 73}{8}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

cos (x) = \frac{3 + 73}{8} :

No Solution

cos (x) = \frac{3 + 73}{8}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

3 cos (x) - 4 sin^{2} (x) = 0 : x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

- (1 - u^{2}) \cdot 4 + 3 u = 0

- (1 - u^{2}) \cdot 4 + 3 u = 0 : u = \frac{- 3 + 73}{8}, u = \frac{- 3 - 73}{8}

- (1 - u^{2}) \cdot 4 + 3 u = 0

Expand

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

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

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

Expand

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

- 4 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = u^{2}

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = 3, c = - 4

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

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

3^{2} - 4 \cdot 4 (- 4) = 73

3^{2} - 4 \cdot 4 (- 4)

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 3^{2} + 64

3^{2} = 9

= 9 + 64

Add the numbers:

9 + 64 = 73

= 73

u_{1, 2} = \frac{- 3 \pm 73}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 3 + 73}{2 \cdot 4}, u_{2} = \frac{- 3 - 73}{2 \cdot 4}

u = \frac{- 3 + 73}{2 \cdot 4} : \frac{- 3 + 73}{8}

\frac{- 3 + 73}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 3 + 73}{8}

u = \frac{- 3 - 73}{2 \cdot 4} : \frac{- 3 - 73}{8}

\frac{- 3 - 73}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 3 - 73}{8}

The solutions to the quadratic equation are:

u = \frac{- 3 + 73}{8}, u = \frac{- 3 - 73}{8}

Substitute back

u = cos (x)

cos (x) = \frac{- 3 + 73}{8}, cos (x) = \frac{- 3 - 73}{8}

cos (x) = \frac{- 3 + 73}{8}, cos (x) = \frac{- 3 - 73}{8}

cos (x) = \frac{- 3 + 73}{8} : x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

cos (x) = \frac{- 3 + 73}{8}

Apply trig inverse properties

cos (x) = \frac{- 3 + 73}{8}

General solutions for

cos (x) = \frac{- 3 + 73}{8}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

cos (x) = \frac{- 3 - 73}{8} :

No Solution

cos (x) = \frac{- 3 - 73}{8}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

Combine all the solutions

x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

arccos (- \frac{- 3 + 73}{8}) + 2 πn :

False

arccos (- \frac{- 3 + 73}{8}) + 2 πn

Plug in

n = 1

arccos (- \frac{- 3 + 73}{8}) + 2 π 1

For

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

plug in

x = arccos (- \frac{- 3 + 73}{8}) + 2 π 1

\frac{3}{tan ( arccos ( - \frac{- 3 + 73}{8} ) + 2 π 1 )} = 2 tan (arccos (- \frac{- 3 + 73}{8}) + 2 π 1) \cdot 2 cos (arccos (- \frac{- 3 + 73}{8}) + 2 π 1)

Refine

- 2.88374 \dots = 2.88374 \dots

\Rightarrow False

Check the solution

- arccos (- \frac{- 3 + 73}{8}) + 2 πn :

False

- arccos (- \frac{- 3 + 73}{8}) + 2 πn

Plug in

n = 1

- arccos (- \frac{- 3 + 73}{8}) + 2 π 1

For

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

plug in

x = - arccos (- \frac{- 3 + 73}{8}) + 2 π 1

\frac{3}{tan ( - arccos ( - \frac{- 3 + 73}{8} ) + 2 π 1 )} = 2 tan (- arccos (- \frac{- 3 + 73}{8}) + 2 π 1) \cdot 2 cos (- arccos (- \frac{- 3 + 73}{8}) + 2 π 1)

Refine

2.88374 \dots = - 2.88374 \dots

\Rightarrow False

Check the solution

arccos (\frac{- 3 + 73}{8}) + 2 πn :

True

arccos (\frac{- 3 + 73}{8}) + 2 πn

Plug in

n = 1

arccos (\frac{- 3 + 73}{8}) + 2 π 1

For

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

plug in

x = arccos (\frac{- 3 + 73}{8}) + 2 π 1

\frac{3}{tan ( arccos ( \frac{- 3 + 73}{8} ) + 2 π 1 )} = 2 tan (arccos (\frac{- 3 + 73}{8}) + 2 π 1) \cdot 2 cos (arccos (\frac{- 3 + 73}{8}) + 2 π 1)

Refine

2.88374 \dots = 2.88374 \dots

\Rightarrow True

Check the solution

2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn :

True

2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{- 3 + 73}{8}) + 2 π 1

For

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

plug in

x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 π 1

\frac{3}{tan ( 2 π - arccos ( \frac{- 3 + 73}{8} ) + 2 π 1 )} = 2 tan (2 π - arccos (\frac{- 3 + 73}{8}) + 2 π 1) \cdot 2 cos (2 π - arccos (\frac{- 3 + 73}{8}) + 2 π 1)

Refine

- 2.88374 \dots = - 2.88374 \dots

\Rightarrow True

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

Show solutions in decimal form

x = 0.80515 \dots + 2 πn, x = 2 π - 0.80515 \dots + 2 πn

Graph

Popular Examples

cos(θ)=-4/5 ,sin(θ)

5sin(4x)=4cos(2x)

arccosh(x)=1

solvefor θ,sin(3θ)= 1/2

0=-0.5sin(x/5)

Frequently Asked Questions (FAQ)

What is the general solution for 3/(tan(x))=(2tan(x))(2cos(x)) ?
The general solution for 3/(tan(x))=(2tan(x))(2cos(x)) is x=0.80515…+2pin,x=2pi-0.80515…+2pin