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

The general solution for 2cot(x)+sec^2(x)=0 is x=(3pi)/4+pin

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Popular Trigonometry >

2cot(x)+sec^2(x)=0

Solution

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

Solution

x = \frac{3 π}{4} + πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Multiply the numbers:

1 \cdot 2 = 2

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

= sec^{2} (x) + \frac{2}{tan ( x )}

Use the Pythagorean identity:

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

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

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

Solve by substitution

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

Let:

tan (x) = u

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

1 + \frac{2}{u} + u^{2} = 0 : u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

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

Multiply both sides by

u

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

Multiply both sides by

u

1 \cdot u + \frac{2}{u} u + u^{2} u = 0 \cdot u

Simplify

1 \cdot u + \frac{2}{u} u + u^{2} u = 0 \cdot u

Simplify

1 \cdot u : u

1 \cdot u

Multiply:

1 \cdot u = u

= u

Simplify

\frac{2}{u} u : 2

\frac{2}{u} u

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 u}{u}

Cancel the common factor:

u

= 2

Simplify

u^{2} u : u^{3}

u^{2} u

Apply exponent rule:

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

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

= u^{2 + 1}

Add the numbers:

2 + 1 = 3

= u^{3}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u + 2 + u^{3} = 0

u + 2 + u^{3} = 0

u + 2 + u^{3} = 0

Solve

u + 2 + u^{3} = 0 : u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

u + 2 + u^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{3} + u + 2 = 0

Factor

u^{3} + u + 2 : (u + 1) (u^{2} - u + 2)

u^{3} + u + 2

Use the rational root theorem

a_{0} = 2, a_{n} = 1

The dividers of

a_{0} : 1, 2,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1 , 2}{1}

- \frac{1}{1}

is a root of the expression, so factor out

u + 1

= (u + 1) \frac{u ^{3} + u + 2}{u + 1}

\frac{u ^{3} + u + 2}{u + 1} = u^{2} - u + 2

\frac{u ^{3} + u + 2}{u + 1}

Divide

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

Divide the leading coefficients of the numerator

u^{3} + u + 2

and the divisor

u + 1 : \frac{u ^{3}}{u} = u^{2}

Quotient = u^{2}

Multiply

u + 1

u^{2} : u^{3} + u^{2}

Subtract

u^{3} + u^{2}

from

u^{3} + u + 2

to get new remainder

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

Therefore

\frac{u ^{3} + u + 2}{u + 1} = u^{2} + \frac{- u ^{2} + u + 2}{u + 1}

= u^{2} + \frac{- u ^{2} + u + 2}{u + 1}

Divide

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

Divide the leading coefficients of the numerator

- u^{2} + u + 2

and the divisor

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

Quotient = - u

Multiply

u + 1

- u : - u^{2} - u

Subtract

- u^{2} - u

from

- u^{2} + u + 2

to get new remainder

Remainder = 2 u + 2

Therefore

\frac{- u ^{2} + u + 2}{u + 1} = - u + \frac{2 u + 2}{u + 1}

= u^{2} - u + \frac{2 u + 2}{u + 1}

Divide

\frac{2 u + 2}{u + 1} : \frac{2 u + 2}{u + 1} = 2

Divide the leading coefficients of the numerator

2 u + 2

and the divisor

u + 1 : \frac{2 u}{u} = 2

Quotient = 2

Multiply

u + 1

2 : 2 u + 2

Subtract

2 u + 2

from

2 u + 2

to get new remainder

Remainder = 0

Therefore

\frac{2 u + 2}{u + 1} = 2

= u^{2} - u + 2

= (u + 1) (u^{2} - u + 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

u^{2} - u + 2 = 0 : u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

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}

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 2 : 7 i

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 - 8

Subtract the numbers:

1 - 8 = - 7

= - 7

Apply radical rule:

- a = - 1 a

- 7 = - 1 7

= - 1 7

Apply imaginary number rule:

- 1 = i

= 7 i

u_{1, 2} = \frac{- ( - 1 ) \pm 7 i}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7 i}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 7 i}{2 \cdot 1}

u = \frac{- ( - 1 ) + 7 i}{2 \cdot 1} : \frac{1}{2} + i \frac{7}{2}

\frac{- ( - 1 ) + 7 i}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 + 7 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 7 i}{2}

Rewrite

\frac{1 + 7 i}{2}

in standard complex form:

\frac{1}{2} + \frac{7}{2} i

\frac{1 + 7 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 + 7 i}{2} = \frac{1}{2} + \frac{7 i}{2}

= \frac{1}{2} + \frac{7 i}{2}

= \frac{1}{2} + \frac{7}{2} i

u = \frac{- ( - 1 ) - 7 i}{2 \cdot 1} : \frac{1}{2} - i \frac{7}{2}

\frac{- ( - 1 ) - 7 i}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 - 7 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 7 i}{2}

Rewrite

\frac{1 - 7 i}{2}

in standard complex form:

\frac{1}{2} - \frac{7}{2} i

\frac{1 - 7 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 - 7 i}{2} = \frac{1}{2} - \frac{7 i}{2}

= \frac{1}{2} - \frac{7 i}{2}

= \frac{1}{2} - \frac{7}{2} i

The solutions to the quadratic equation are:

u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

The solutions are

u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

1 + \frac{2}{u} + u^{2}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

Substitute back

u = tan (x)

tan (x) = - 1, tan (x) = \frac{1}{2} + i \frac{7}{2}, tan (x) = \frac{1}{2} - i \frac{7}{2}

tan (x) = - 1, tan (x) = \frac{1}{2} + i \frac{7}{2}, tan (x) = \frac{1}{2} - i \frac{7}{2}

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

tan (x) = \frac{1}{2} + i \frac{7}{2} :

No Solution

tan (x) = \frac{1}{2} + i \frac{7}{2}

No Solution

tan (x) = \frac{1}{2} - i \frac{7}{2} :

No Solution

tan (x) = \frac{1}{2} - i \frac{7}{2}

No Solution

Combine all the solutions

x = \frac{3 π}{4} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cot(x)+sec^2(x)=0 ?
The general solution for 2cot(x)+sec^2(x)=0 is x=(3pi)/4+pin