What is the general solution for tan(2x)=-tan(x)2 ?

The general solution for tan(2x)=-tan(x)2 is x=pin,x=-0.95531…+pin,x=0.95531…+pin

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Popular >

tan(2x)=-tan(x)2

Solution

tan (2 x) = - tan (x) 2

Solution

x = πn, x = - 0.95531 \dots + πn, x = 0.95531 \dots + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = - 54.73561 \dots^{\circ} + 18 0^{\circ} n, x = 54.73561 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan (2 x) = - tan (x) \cdot 2

Subtract

- tan (x) 2

from both sides

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

Rewrite using trig identities

tan (2 x) + 2 tan (x)

Use the Double Angle identity:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

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

Simplify

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

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} + 2 tan (x)

Convert element to fraction:

2 tan (x) = \frac{2 tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Expand

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

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

Expand

2 tan (x) (1 - tan^{2} (x)) : 2 tan (x) - 2 tan^{3} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2 tan (x), b = 1, c = tan^{2} (x)

= 2 tan (x) \cdot 1 - 2 tan (x) tan^{2} (x)

= 2 \cdot 1 \cdot tan (x) - 2 tan^{2} (x) tan (x)

Simplify

2 \cdot 1 \cdot tan (x) - 2 tan^{2} (x) tan (x) : 2 tan (x) - 2 tan^{3} (x)

2 \cdot 1 \cdot tan (x) - 2 tan^{2} (x) tan (x)

2 \cdot 1 \cdot tan (x) = 2 tan (x)

2 \cdot 1 \cdot tan (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 tan (x)

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

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

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 tan^{3} (x)

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

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

= 2 tan (x) + 2 tan (x) - 2 tan^{3} (x)

Add similar elements:

2 tan (x) + 2 tan (x) = 4 tan (x)

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

= \frac{4 tan ( x ) - 2 tan ^{3} ( x )}{1 - tan ^{2} ( x )}

= \frac{4 tan ( x ) - 2 tan ^{3} ( x )}{1 - tan ^{2} ( x )}

\frac{- 2 tan ^{3} ( x ) + 4 tan ( x )}{1 - tan ^{2} ( x )} = 0

Solve by substitution

\frac{- 2 tan ^{3} ( x ) + 4 tan ( x )}{1 - tan ^{2} ( x )} = 0

Let:

tan (x) = u

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

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

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

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

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

Solve

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

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

Factor

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

- 2 u^{3} + 4 u

Factor out common term

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

- 2 u^{3} + 4 u

Apply exponent rule:

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

u^{3} = u^{2} u

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

Rewrite

4

2 \cdot 2

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

Factor out common term

- 2 u

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

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

Factor

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

u^{2} - 2

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (u + 2) (u - 2)

= - 2 u (u + 2) (u - 2)

- 2 u (u + 2) (u - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u + 2 = 0 or u - 2 = 0

Solve

u + 2 = 0 : u = - 2

u + 2 = 0

Move

2

to the right side

u + 2 = 0

Subtract

2

from both sides

u + 2 - 2 = 0 - 2

Simplify

u = - 2

u = - 2

Solve

u - 2 = 0 : u = 2

u - 2 = 0

Move

2

to the right side

u - 2 = 0

Add

2

to both sides

u - 2 + 2 = 0 + 2

Simplify

u = 2

u = 2

The solutions are

u = 0, u = - 2, u = 2

u = 0, u = - 2, u = 2

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

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

and compare to zero

Solve

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

1 - u^{2} = 0

Move

1

to the right side

1 - u^{2} = 0

Subtract

1

from both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The following points are undefined

u = 1, u = - 1

Combine undefined points with solutions:

u = 0, u = - 2, u = 2

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = - 2, tan (x) = 2

tan (x) = 0, tan (x) = - 2, tan (x) = 2

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = - 2 : x = arctan (- 2) + πn

tan (x) = - 2

Apply trig inverse properties

tan (x) = - 2

General solutions for

tan (x) = - 2

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- 2) + πn

x = arctan (- 2) + πn

tan (x) = 2 : x = arctan (2) + πn

tan (x) = 2

Apply trig inverse properties

tan (x) = 2

General solutions for

tan (x) = 2

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (2) + πn

x = arctan (2) + πn

Combine all the solutions

x = πn, x = arctan (- 2) + πn, x = arctan (2) + πn

Show solutions in decimal form

x = πn, x = - 0.95531 \dots + πn, x = 0.95531 \dots + πn

Popular Examples

2sin^2(x)-5sin(x)-12=0

2 sin^{2} (x) - 5 sin (x) - 12 = 0

1/(sin(x))=sqrt(5)

\frac{1}{sin ( x )} = 5

tan(2x)+tan(4x)=0

tan (2 x) + tan (4 x) = 0

2sin(θ)-sqrt(3)=0,0<= θ<2pi

2 sin (θ) - 3 = 0, 0 \leq θ < 2 π

arctan(θ)=-1

arctan (θ) = - 1

Frequently Asked Questions (FAQ)

What is the general solution for tan(2x)=-tan(x)2 ?
The general solution for tan(2x)=-tan(x)2 is x=pin,x=-0.95531…+pin,x=0.95531…+pin