What is the general solution for tan^2(x)=2tan(x)+2tan^3(x) ?

The general solution for tan^2(x)=2tan(x)+2tan^3(x) is x=pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(x)=2tan(x)+2tan^3(x)

Solution

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

Solution

x = πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

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

u^{2} = 2 u + 2 u^{3} : u = 0, u = \frac{1}{4} + i \frac{15}{4}, u = \frac{1}{4} - i \frac{15}{4}

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

Switch sides

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

Move

u^{2}

to the left side

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

Subtract

u^{2}

from both sides

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

Simplify

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

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

Factor

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

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

Apply exponent rule:

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

u^{2} = uu

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

Factor out common term

u

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 u^{2} - u + 2 = 0 : u = \frac{1}{4} + i \frac{15}{4}, u = \frac{1}{4} - i \frac{15}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 2 : 15 i

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 - 16

Subtract the numbers:

1 - 16 = - 15

= - 15

Apply radical rule:

- a = - 1 a

- 15 = - 1 15

= - 1 15

Apply imaginary number rule:

- 1 = i

= 15 i

u_{1, 2} = \frac{- ( - 1 ) \pm 15 i}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 15 i}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 15 i}{2 \cdot 2}

u = \frac{- ( - 1 ) + 15 i}{2 \cdot 2} : \frac{1}{4} + i \frac{15}{4}

\frac{- ( - 1 ) + 15 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 15 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 15 i}{4}

Rewrite

\frac{1 + 15 i}{4}

in standard complex form:

\frac{1}{4} + \frac{15}{4} i

\frac{1 + 15 i}{4}

Apply the fraction rule:

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

\frac{1 + 15 i}{4} = \frac{1}{4} + \frac{15 i}{4}

= \frac{1}{4} + \frac{15 i}{4}

= \frac{1}{4} + \frac{15}{4} i

u = \frac{- ( - 1 ) - 15 i}{2 \cdot 2} : \frac{1}{4} - i \frac{15}{4}

\frac{- ( - 1 ) - 15 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 15 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 15 i}{4}

Rewrite

\frac{1 - 15 i}{4}

in standard complex form:

\frac{1}{4} - \frac{15}{4} i

\frac{1 - 15 i}{4}

Apply the fraction rule:

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

\frac{1 - 15 i}{4} = \frac{1}{4} - \frac{15 i}{4}

= \frac{1}{4} - \frac{15 i}{4}

= \frac{1}{4} - \frac{15}{4} i

The solutions to the quadratic equation are:

u = \frac{1}{4} + i \frac{15}{4}, u = \frac{1}{4} - i \frac{15}{4}

The solutions are

u = 0, u = \frac{1}{4} + i \frac{15}{4}, u = \frac{1}{4} - i \frac{15}{4}

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = \frac{1}{4} + i \frac{15}{4}, tan (x) = \frac{1}{4} - i \frac{15}{4}

tan (x) = 0, tan (x) = \frac{1}{4} + i \frac{15}{4}, tan (x) = \frac{1}{4} - i \frac{15}{4}

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) = \frac{1}{4} + i \frac{15}{4} :

No Solution

tan (x) = \frac{1}{4} + i \frac{15}{4}

No Solution

tan (x) = \frac{1}{4} - i \frac{15}{4} :

No Solution

tan (x) = \frac{1}{4} - i \frac{15}{4}

No Solution

Combine all the solutions

x = πn

Popular Examples

solvefor t,cos(wt+k)=(20)/((2*pi*f)^2)solve for

t, cos (wt + k) = \frac{20}{( 2 \cdot π \cdot f ) ^{2}}

0=-cos(x)(2sin(x)+3)

0 = - cos (x) (2 sin (x) + 3)

28cos(t)=20,0<= t<360

28 cos (t) = 20, 0^{\circ} \leq t < 36 0^{\circ}

8tan(x/2)+8cos(x)tan(x/2)=1

8 tan (\frac{x}{2}) + 8 cos (x) tan (\frac{x}{2}) = 1

sin(a)=0.5937

sin (a) = 0.5937

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)=2tan(x)+2tan^3(x) ?
The general solution for tan^2(x)=2tan(x)+2tan^3(x) is x=pin