What is the general solution for 2sec^2(x)+tan(x)-3=0 ?

The general solution for 2sec^2(x)+tan(x)-3=0 is x=0.46364…+pin,x=(3pi)/4+pin

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

Solution

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

Solution

x = 0.46364 \dots + πn, x = \frac{3 π}{4} + πn

Degrees

x = 26.56505 \dots^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 3 + 2 = - 1

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

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

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

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 1

Substitute back

u = tan (x)

tan (x) = \frac{1}{2}, tan (x) = - 1

tan (x) = \frac{1}{2}, tan (x) = - 1

tan (x) = \frac{1}{2} : x = arctan (\frac{1}{2}) + πn

tan (x) = \frac{1}{2}

Apply trig inverse properties

tan (x) = \frac{1}{2}

General solutions for

tan (x) = \frac{1}{2}

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

x = arctan (\frac{1}{2}) + πn

x = arctan (\frac{1}{2}) + πn

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = arctan (\frac{1}{2}) + πn, x = \frac{3 π}{4} + πn

Show solutions in decimal form

x = 0.46364 \dots + πn, x = \frac{3 π}{4} + πn

Popular Examples

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8 cos (θ) = 2 sec (θ)

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Frequently Asked Questions (FAQ)

What is the general solution for 2sec^2(x)+tan(x)-3=0 ?
The general solution for 2sec^2(x)+tan(x)-3=0 is x=0.46364…+pin,x=(3pi)/4+pin