What is the general solution for 2tan(x)=3-tan^2(x),-2pi<= ,x<= pi ?

Q: What is the general solution for 2tan(x)=3-tan^2(x),-2pi<= ,x<= pi ?

The general solution for 2tan(x)=3-tan^2(x),-2pi<= ,x<= pi is No Solution for x\in\mathbb{R}

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2tan(x)=3-tan^2(x),-2pi<= ,x<= pi

Solution

2 tan (x) = 3 - tan^{2} (x), - 2 π \leq, x \leq π

Solution

No Solution for x \in R

Solution steps

2 tan (x) = 3 - tan^{2} (x), - 2 π, x \leq π

Solve by substitution

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

Let:

tan (x) = u

2 u = 3 - u^{2}

2 u = 3 - u^{2} : u = - 3, u = 1

2 u = 3 - u^{2}

Switch sides

3 - u^{2} = 2 u

Move

2 u

to the left side

3 - u^{2} = 2 u

Subtract

2 u

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = - 2, c = 3

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

u = \frac{- ( - 2 ) + 4}{2 ( - 1 )} : - 3

\frac{- ( - 2 ) + 4}{2 ( - 1 )}

Remove parentheses:

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

= \frac{2 + 4}{- 2 \cdot 1}

Add the numbers:

2 + 4 = 6

= \frac{6}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{- 2}

Apply the fraction rule:

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

= - \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3

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

\frac{- ( - 2 ) - 4}{2 ( - 1 )}

Remove parentheses:

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

= \frac{2 - 4}{- 2 \cdot 1}

Subtract the numbers:

2 - 4 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

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

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - 3, u = 1

Substitute back

u = tan (x)

tan (x) = - 3, tan (x) = 1

tan (x) = - 3, tan (x) = 1

tan (x) = - 3, - 2 π :

No Solution

tan (x) = - 3, - 2 π

Apply trig inverse properties

tan (x) = - 3

General solutions for

tan (x) = - 3

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

x = arctan (- 3) + πn

x = arctan (- 3) + πn

Solutions for the range

- 2 π

No Solution

tan (x) = 1, - 2 π :

No Solution

tan (x) = 1, - 2 π

General solutions for

tan (x) = 1

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 = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solutions for the range

- 2 π

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan(x)=3-tan^2(x),-2pi<= ,x<= pi ?
The general solution for 2tan(x)=3-tan^2(x),-2pi<= ,x<= pi is No Solution for x\in\mathbb{R}