What is the general solution for 2tan(x)+cos(x)=0 ?

The general solution for 2tan(x)+cos(x)=0 is x=-0.42707…+2pin,x=pi+0.42707…+2pin

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Popular Trigonometry >

2tan(x)+cos(x)=0

Solution

2 tan (x) + cos (x) = 0

Solution

x = - 0.42707 \dots + 2 πn, x = π + 0.42707 \dots + 2 πn

Degrees

x = - 24.46980 \dots^{\circ} + 36 0^{\circ} n, x = 204.46980 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 tan (x) + cos (x) = 0

Express with sin, cos

2 \cdot \frac{sin ( x )}{cos ( x )} + cos (x) = 0

Simplify

2 \cdot \frac{sin ( x )}{cos ( x )} + cos (x) : \frac{2 sin ( x ) + cos ^{2} ( x )}{cos ( x )}

2 \cdot \frac{sin ( x )}{cos ( x )} + cos (x)

Multiply

2 \cdot \frac{sin ( x )}{cos ( x )} : \frac{2 sin ( x )}{cos ( x )}

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

Multiply fractions:

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (x) \cdot 2 + cos (x) cos (x)

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

Subtract

cos^{2} (x)

from both sides

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

Square both sides

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

Subtract

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

from both sides

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

Factor

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

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

Rewrite

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

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

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

Rewrite

4

2^{2}

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

Solving each part separately

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

2 sin (x) + cos^{2} (x) = 0 : x = arcsin (1 - 2) + 2 πn, x = π + arcsin (- 1 + 2) + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 (- 1) \cdot 1 = 22

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

Cancel

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

\frac{- 2 + 2 2}{2}

Factor

- 2 + 22 : 2 (- 1 + 2)

- 2 + 22

Rewrite as

= - 2 \cdot 1 + 22

Factor out common term

2

= 2 (- 1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 2

= - (2 - 1)

Distribute parentheses

= - (- 1) - (2)

Apply minus-plus rules

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

= 1 - 2

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 - 2 2}{- 2}

Apply the fraction rule:

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

- 2 - 22 = - (2 + 22)

= \frac{2 + 2 2}{2}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

The solutions to the quadratic equation are:

u = 1 - 2, u = 1 + 2

Substitute back

u = sin (x)

sin (x) = 1 - 2, sin (x) = 1 + 2

sin (x) = 1 - 2, sin (x) = 1 + 2

sin (x) = 1 - 2 : x = arcsin (1 - 2) + 2 πn, x = π + arcsin (- 1 + 2) + 2 πn

sin (x) = 1 - 2

Apply trig inverse properties

sin (x) = 1 - 2

General solutions for

sin (x) = 1 - 2

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (1 - 2) + 2 πn, x = π + arcsin (- 1 + 2) + 2 πn

x = arcsin (1 - 2) + 2 πn, x = π + arcsin (- 1 + 2) + 2 πn

sin (x) = 1 + 2 :

No Solution

sin (x) = 1 + 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (1 - 2) + 2 πn, x = π + arcsin (- 1 + 2) + 2 πn

2 sin (x) - cos^{2} (x) = 0 : x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 2 + 22 : 2 (- 1 + 2)

- 2 + 22

Rewrite as

= - 2 \cdot 1 + 22

Factor out common term

2

= 2 (- 1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 2

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 - 2 2}{2}

Factor

- 2 - 22 : - 2 (1 + 2)

- 2 - 22

Rewrite as

= - 2 \cdot 1 - 22

Factor out common term

2

= - 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= - (1 + 2)

Negate

- (1 + 2) = - 1 - 2

= - 1 - 2

The solutions to the quadratic equation are:

u = - 1 + 2, u = - 1 - 2

Substitute back

u = sin (x)

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2 : x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 + 2

Apply trig inverse properties

sin (x) = - 1 + 2

General solutions for

sin (x) = - 1 + 2

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 - 2 :

No Solution

sin (x) = - 1 - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

Combine all the solutions

x = arcsin (1 - 2) + 2 πn, x = π + arcsin (- 1 + 2) + 2 πn, x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 tan (x) + cos (x) = 0

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

Check the solution

arcsin (1 - 2) + 2 πn :

True

arcsin (1 - 2) + 2 πn

Plug in

n = 1

arcsin (1 - 2) + 2 π 1

For

2 tan (x) + cos (x) = 0

plug in

x = arcsin (1 - 2) + 2 π 1

2 tan (arcsin (1 - 2) + 2 π 1) + cos (arcsin (1 - 2) + 2 π 1) = 0

Refine

0 = 0

\Rightarrow True

Check the solution

π + arcsin (- 1 + 2) + 2 πn :

True

π + arcsin (- 1 + 2) + 2 πn

Plug in

n = 1

π + arcsin (- 1 + 2) + 2 π 1

For

2 tan (x) + cos (x) = 0

plug in

x = π + arcsin (- 1 + 2) + 2 π 1

2 tan (π + arcsin (- 1 + 2) + 2 π 1) + cos (π + arcsin (- 1 + 2) + 2 π 1) = 0

Refine

0 = 0

\Rightarrow True

Check the solution

arcsin (- 1 + 2) + 2 πn :

False

arcsin (- 1 + 2) + 2 πn

Plug in

n = 1

arcsin (- 1 + 2) + 2 π 1

For

2 tan (x) + cos (x) = 0

plug in

x = arcsin (- 1 + 2) + 2 π 1

2 tan (arcsin (- 1 + 2) + 2 π 1) + cos (arcsin (- 1 + 2) + 2 π 1) = 0

Refine

1.82035 \dots = 0

\Rightarrow False

Check the solution

π - arcsin (- 1 + 2) + 2 πn :

False

π - arcsin (- 1 + 2) + 2 πn

Plug in

n = 1

π - arcsin (- 1 + 2) + 2 π 1

For

2 tan (x) + cos (x) = 0

plug in

x = π - arcsin (- 1 + 2) + 2 π 1

2 tan (π - arcsin (- 1 + 2) + 2 π 1) + cos (π - arcsin (- 1 + 2) + 2 π 1) = 0

Refine

- 1.82035 \dots = 0

\Rightarrow False

x = arcsin (1 - 2) + 2 πn, x = π + arcsin (- 1 + 2) + 2 πn

Show solutions in decimal form

x = - 0.42707 \dots + 2 πn, x = π + 0.42707 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan(x)+cos(x)=0 ?
The general solution for 2tan(x)+cos(x)=0 is x=-0.42707…+2pin,x=pi+0.42707…+2pin