What is the general solution for 4tan^2(x)+sec(x)=-2tan^2(x)+sec^2(x)-3 ?

The general solution for 4tan^2(x)+sec(x)=-2tan^2(x)+sec^2(x)-3 is No Solution for x\in\mathbb{R}

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4tan^2(x)+sec(x)=-2tan^2(x)+sec^2(x)-3

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

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

from both sides

6 tan^{2} (x) + sec (x) - sec^{2} (x) + 3 = 0

Rewrite using trig identities

3 + sec (x) - sec^{2} (x) + 6 tan^{2} (x)

Use the Pythagorean identity:

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

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

= 3 + sec (x) - sec^{2} (x) + 6 (sec^{2} (x) - 1)

Simplify

3 + sec (x) - sec^{2} (x) + 6 (sec^{2} (x) - 1) : 5 sec^{2} (x) + sec (x) - 3

3 + sec (x) - sec^{2} (x) + 6 (sec^{2} (x) - 1)

Expand

6 (sec^{2} (x) - 1) : 6 sec^{2} (x) - 6

6 (sec^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = sec^{2} (x), c = 1

= 6 sec^{2} (x) - 6 \cdot 1

Multiply the numbers:

6 \cdot 1 = 6

= 6 sec^{2} (x) - 6

= 3 + sec (x) - sec^{2} (x) + 6 sec^{2} (x) - 6

Simplify

3 + sec (x) - sec^{2} (x) + 6 sec^{2} (x) - 6 : 5 sec^{2} (x) + sec (x) - 3

3 + sec (x) - sec^{2} (x) + 6 sec^{2} (x) - 6

Add similar elements:

- sec^{2} (x) + 6 sec^{2} (x) = 5 sec^{2} (x)

= 3 + sec (x) + 5 sec^{2} (x) - 6

Group like terms

= sec (x) + 5 sec^{2} (x) + 3 - 6

Add/Subtract the numbers:

3 - 6 = - 3

= 5 sec^{2} (x) + sec (x) - 3

= 5 sec^{2} (x) + sec (x) - 3

= 5 sec^{2} (x) + sec (x) - 3

- 3 + sec (x) + 5 sec^{2} (x) = 0

Solve by substitution

- 3 + sec (x) + 5 sec^{2} (x) = 0

Let:

sec (x) = u

- 3 + u + 5 u^{2} = 0

- 3 + u + 5 u^{2} = 0 : u = \frac{- 1 + 61}{10}, u = \frac{- 1 - 61}{10}

- 3 + u + 5 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 5, b = 1, c = - 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

1^{2} - 4 \cdot 5 (- 3) = 61

1^{2} - 4 \cdot 5 (- 3)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 5 (- 3)

Apply rule

- (- a) = a

= 1 + 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 1 + 60

Add the numbers:

1 + 60 = 61

= 61

u_{1, 2} = \frac{- 1 \pm 61}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 1 + 61}{2 \cdot 5}, u_{2} = \frac{- 1 - 61}{2 \cdot 5}

u = \frac{- 1 + 61}{2 \cdot 5} : \frac{- 1 + 61}{10}

\frac{- 1 + 61}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 1 + 61}{10}

u = \frac{- 1 - 61}{2 \cdot 5} : \frac{- 1 - 61}{10}

\frac{- 1 - 61}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 1 - 61}{10}

The solutions to the quadratic equation are:

u = \frac{- 1 + 61}{10}, u = \frac{- 1 - 61}{10}

Substitute back

u = sec (x)

sec (x) = \frac{- 1 + 61}{10}, sec (x) = \frac{- 1 - 61}{10}

sec (x) = \frac{- 1 + 61}{10}, sec (x) = \frac{- 1 - 61}{10}

sec (x) = \frac{- 1 + 61}{10} :

No Solution

sec (x) = \frac{- 1 + 61}{10}

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

sec (x) = \frac{- 1 - 61}{10} :

No Solution

sec (x) = \frac{- 1 - 61}{10}

sec (x) \leq - 1 or sec (x) \geq 1

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 4tan^2(x)+sec(x)=-2tan^2(x)+sec^2(x)-3 ?
The general solution for 4tan^2(x)+sec(x)=-2tan^2(x)+sec^2(x)-3 is No Solution for x\in\mathbb{R}