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

The general solution for sec^2(x)+tan(x)=3 is x= pi/4+pin,x=-1.10714…+pin

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

Solution

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

Solution

x = \frac{π}{4} + πn, x = - 1.10714 \dots + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

3

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

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

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Write in the standard form

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

u^{2} + u - 2 = 0

Solve with the quadratic formula

u^{2} + u - 2 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 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 1}

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = 1, u = - 2

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = - 2

tan (x) = 1, tan (x) = - 2

tan (x) = 1 : x = \frac{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{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 2 : x = arctan (- 2) + πn

tan (x) = - 2

Apply trig inverse properties

tan (x) = - 2

General solutions for

tan (x) = - 2

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

x = arctan (- 2) + πn

x = arctan (- 2) + πn

Combine all the solutions

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

Show solutions in decimal form

x = \frac{π}{4} + πn, x = - 1.10714 \dots + πn

Popular Examples

4sin^2(x)-4sin(x)+1=0

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

tan(x)=5

tan (x) = 5

2sin^3(x)=sin(x)

2 sin^{3} (x) = sin (x)

3cos^2(x)+sin^2(x)=2

3 cos^{2} (x) + sin^{2} (x) = 2

sinh(x)=0

sinh (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)+tan(x)=3 ?
The general solution for sec^2(x)+tan(x)=3 is x= pi/4+pin,x=-1.10714…+pin