What is the general solution for tan^2(x)+3sec(x)+3=0,0<= x<= 360 ?

Q: What is the general solution for tan^2(x)+3sec(x)+3=0,0<= x<= 360 ?

The general solution for tan^2(x)+3sec(x)+3=0,0<= x<= 360 is x=180,x=120,x=240

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tan^2(x)+3sec(x)+3=0,0<= x<= 360

Solution

tan^{2} (x) + 3 sec (x) + 3 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 18 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

Radians

x = π, x = \frac{2 π}{3}, x = \frac{4 π}{3}

Solution steps

tan^{2} (x) + 3 sec (x) + 3 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

3 - 1 = 2

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

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

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

Solve by substitution

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

Let:

sec (x) = u

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

2 + u^{2} + 3 u = 0 : u = - 1, u = - 2

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 3, c = 2

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

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

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 3 + 1 = - 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

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

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

Subtract the numbers:

- 3 - 1 = - 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 = sec (x)

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

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

sec (x) = - 1, 0 \leq x \leq 36 0^{\circ} : x = 18 0^{\circ}

sec (x) = - 1, 0 \leq x \leq 36 0^{\circ}

General solutions for

sec (x) = - 1

sec (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 18 0^{\circ} + 36 0^{\circ} n

x = 18 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 18 0^{\circ}

sec (x) = - 2, 0 \leq x \leq 36 0^{\circ} : x = 12 0^{\circ}, x = 24 0^{\circ}

sec (x) = - 2, 0 \leq x \leq 36 0^{\circ}

General solutions for

sec (x) = - 2

sec (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 12 0^{\circ}, x = 24 0^{\circ}

Combine all the solutions

x = 18 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+3sec(x)+3=0,0<= x<= 360 ?
The general solution for tan^2(x)+3sec(x)+3=0,0<= x<= 360 is x=180,x=120,x=240