What is the general solution for 2tan(x)+sec^2(x)=0,-90<x<90 ?

Q: What is the general solution for 2tan(x)+sec^2(x)=0,-90<x<90 ?

The general solution for 2tan(x)+sec^2(x)=0,-90<x<90 is x=-45

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2tan(x)+sec^2(x)=0,-90<x<90

Solution

2 tan (x) + sec^{2} (x) = 0, - 9 0^{\circ} < x < 9 0^{\circ}

Solution

x = - 4 5^{\circ}

Radians

x = - \frac{π}{4}

Solution steps

2 tan (x) + sec^{2} (x) = 0, - 9 0^{\circ} < x < 9 0^{\circ}

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 2, c = 1

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

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

2^{2} - 4 \cdot 1 \cdot 1 = 0

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

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

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

\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 = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = tan (x)

tan (x) = - 1

tan (x) = - 1

tan (x) = - 1, - 9 0^{\circ} < x < 9 0^{\circ} : x = - 4 5^{\circ}

tan (x) = - 1, - 9 0^{\circ} < x < 9 0^{\circ}

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

18 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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

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

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

Solutions for the range

- 9 0^{\circ} < x < 9 0^{\circ}

x = - 4 5^{\circ}

Combine all the solutions

x = - 4 5^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan(x)+sec^2(x)=0,-90<x<90 ?
The general solution for 2tan(x)+sec^2(x)=0,-90<x<90 is x=-45