What is the general solution for 2sec^2(x)+4tan(x)=1 ?

The general solution for 2sec^2(x)+4tan(x)=1 is x=-0.28492…+pin,x=-1.04089…+pin

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2sec^2(x)+4tan(x)=1

Solution

2 sec^{2} (x) + 4 tan (x) = 1

Solution

x = - 0.28492 \dots + πn, x = - 1.04089 \dots + πn

Degrees

x = - 16.32494 \dots^{\circ} + 18 0^{\circ} n, x = - 59.63880 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 sec^{2} (x) + 4 tan (x) = 1

Subtract

1

from both sides

2 sec^{2} (x) + 4 tan (x) - 1 = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b + c) = ab + a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 2 = 1

= 2 tan^{2} (x) + 4 tan (x) + 1

= 2 tan^{2} (x) + 4 tan (x) + 1

= 2 tan^{2} (x) + 4 tan (x) + 1

1 + 2 tan^{2} (x) + 4 tan (x) = 0

Solve by substitution

1 + 2 tan^{2} (x) + 4 tan (x) = 0

Let:

tan (x) = u

1 + 2 u^{2} + 4 u = 0

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

1 + 2 u^{2} + 4 u = 0

Write in the standard form

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

2 u^{2} + 4 u + 1 = 0

Solve with the quadratic formula

2 u^{2} + 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 4, c = 1

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

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

4^{2} - 4 \cdot 2 \cdot 1 = 22

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 4^{2} - 8

4^{2} = 16

= 16 - 8

Subtract the numbers:

16 - 8 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

u = \frac{- 4 + 2 2}{2 \cdot 2} : \frac{- 2 + 2}{2}

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

Multiply the numbers:

2 \cdot 2 = 4

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

Factor

- 4 + 22 : 2 (- 2 + 2)

- 4 + 22

Rewrite as

= - 2 \cdot 2 + 22

Factor out common term

2

= 2 (- 2 + 2)

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

Cancel the common factor:

2

= \frac{- 2 + 2}{2}

u = \frac{- 4 - 2 2}{2 \cdot 2} : - \frac{2 + 2}{2}

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

Multiply the numbers:

2 \cdot 2 = 4

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

Factor

- 4 - 22 : - 2 (2 + 2)

- 4 - 22

Rewrite as

= - 2 \cdot 2 - 22

Factor out common term

2

= - 2 (2 + 2)

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

Cancel the common factor:

2

= - \frac{2 + 2}{2}

The solutions to the quadratic equation are:

u = \frac{- 2 + 2}{2}, u = - \frac{2 + 2}{2}

Substitute back

u = tan (x)

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

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

tan (x) = \frac{- 2 + 2}{2} : x = arctan (\frac{- 2 + 2}{2}) + πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (\frac{- 2 + 2}{2}) + πn

x = arctan (\frac{- 2 + 2}{2}) + πn

tan (x) = - \frac{2 + 2}{2} : x = arctan (- \frac{2 + 2}{2}) + πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (- \frac{2 + 2}{2}) + πn

x = arctan (- \frac{2 + 2}{2}) + πn

Combine all the solutions

x = arctan (\frac{- 2 + 2}{2}) + πn, x = arctan (- \frac{2 + 2}{2}) + πn

Show solutions in decimal form

x = - 0.28492 \dots + πn, x = - 1.04089 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sec^2(x)+4tan(x)=1 ?
The general solution for 2sec^2(x)+4tan(x)=1 is x=-0.28492…+pin,x=-1.04089…+pin