What is the general solution for 0=sec^2(x)e^{tan(x)}(2tan(x)+sec^2(x)) ?

The general solution for 0=sec^2(x)e^{tan(x)}(2tan(x)+sec^2(x)) is x=(3pi)/4+pin

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

Solution

0 = sec^{2} (x) e^{t a n (x)} (2 tan (x) + sec^{2} (x))

Solution

x = \frac{3 π}{4} + πn

Degrees

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

Solution steps

0 = sec^{2} (x) e^{t a n (x)} (2 tan (x) + sec^{2} (x))

Switch sides

sec^{2} (x) e^{t a n (x)} (2 tan (x) + sec^{2} (x)) = 0

Rewrite using trig identities

(sec^{2} (x) + 2 tan (x)) sec^{2} (x) e^{t a n (x)}

Use the Pythagorean identity:

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

= (tan^{2} (x) + 1 + 2 tan (x)) (tan^{2} (x) + 1) e^{t a n (x)}

(1 + tan^{2} (x)) (1 + tan^{2} (x) + 2 tan (x)) e^{t a n (x)} = 0

Solve by substitution

(1 + tan^{2} (x)) (1 + tan^{2} (x) + 2 tan (x)) e^{t a n (x)} = 0

Let:

tan (x) = u

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

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

1 + u^{2} = 0 :

No Solution for

u \in R

1 + u^{2} = 0

Move

1

to the right side

1 + u^{2} = 0

Subtract

1

from both sides

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

Simplify

u^{2} = - 1

u^{2} = - 1

x^{2}

cannot be negative for

x \in R

No Solution for u \in R

Solve

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

Solve

e^{u} = 0 :

No Solution for

u \in R

e^{u} = 0

a^{f (u)}

cannot be zero or negative for

u \in R

No Solution for u \in R

u = - 1

Substitute back

u = tan (x)

tan (x) = - 1

tan (x) = - 1

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 = \frac{3 π}{4} + πn

Popular Examples

tan(x)=sqrt(3),-180<x<180

tan (x) = 3, - 18 0^{\circ} < x < 18 0^{\circ}

cot^2(t)csc(t)=sin(t)

cot^{2} (t) csc (t) = sin (t)

6/12 =cos(θ)

\frac{6}{12} = cos (θ)

sin(a)=(6.88)/(6.9)

sin (a) = \frac{6.88}{6.9}

sin(x)=cos(32)

sin (x) = cos (3 2^{\circ})

Frequently Asked Questions (FAQ)

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