What is the general solution for 5tan^2(t)-tan(t)=0 ?

The general solution for 5tan^2(t)-tan(t)=0 is t=0.19739…+pin,t=pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5tan^2(t)-tan(t)=0

Solution

5 tan^{2} (t) - tan (t) = 0

Solution

t = 0.19739 \dots + πn, t = πn

Degrees

t = 11.30993 \dots^{\circ} + 18 0^{\circ} n, t = 0^{\circ} + 18 0^{\circ} n

Solution steps

5 tan^{2} (t) - tan (t) = 0

Solve by substitution

5 tan^{2} (t) - tan (t) = 0

Let:

tan (t) = u

5 u^{2} - u = 0

5 u^{2} - u = 0 : u = \frac{1}{5}, u = 0

5 u^{2} - u = 0

Solve with the quadratic formula

5 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 5, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

(- 1)^{2} - 4 \cdot 5 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 5 \cdot 0

(- 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 5 \cdot 0 = 0

4 \cdot 5 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 1 ) \pm 1}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 1}{2 \cdot 5}, u_{2} = \frac{- ( - 1 ) - 1}{2 \cdot 5}

u = \frac{- ( - 1 ) + 1}{2 \cdot 5} : \frac{1}{5}

\frac{- ( - 1 ) + 1}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{1 + 1}{2 \cdot 5}

Add the numbers:

1 + 1 = 2

= \frac{2}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{2}{10}

Cancel the common factor:

2

= \frac{1}{5}

u = \frac{- ( - 1 ) - 1}{2 \cdot 5} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{1 - 1}{2 \cdot 5}

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{0}{10}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = \frac{1}{5}, u = 0

Substitute back

u = tan (t)

tan (t) = \frac{1}{5}, tan (t) = 0

tan (t) = \frac{1}{5}, tan (t) = 0

tan (t) = \frac{1}{5} : t = arctan (\frac{1}{5}) + πn

tan (t) = \frac{1}{5}

Apply trig inverse properties

tan (t) = \frac{1}{5}

General solutions for

tan (t) = \frac{1}{5}

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

t = arctan (\frac{1}{5}) + πn

t = arctan (\frac{1}{5}) + πn

tan (t) = 0 : t = πn

tan (t) = 0

General solutions for

tan (t) = 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}

t = 0 + πn

t = 0 + πn

Solve

t = 0 + πn : t = πn

t = 0 + πn

0 + πn = πn

t = πn

t = πn

Combine all the solutions

t = arctan (\frac{1}{5}) + πn, t = πn

Show solutions in decimal form

t = 0.19739 \dots + πn, t = πn

Popular Examples

sin(x)= 9/16

sin (x) = \frac{9}{16}

24=32+8-2*sqrt(32*8)*cos(θ)

24 = 32 + 8 - 2 \cdot 32 \cdot 8 \cdot cos (θ)

sin(45+a)cos(45+a)=(sqrt(2))/2

sin (4 5^{\circ} + a) cos (4 5^{\circ} + a) = \frac{2}{2}

tan(x)*sin(x)=5219

tan (x) \cdot sin (x) = 5219

sin(z)=sqrt(2)

sin (z) = 2

Frequently Asked Questions (FAQ)

What is the general solution for 5tan^2(t)-tan(t)=0 ?
The general solution for 5tan^2(t)-tan(t)=0 is t=0.19739…+pin,t=pin