What is the general solution for tan^4(x)-2sec^2(x)+3=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^4(x)-2sec^2(x)+3=0

Solution

tan^{4} (x) - 2 sec^{2} (x) + 3 = 0

Solution

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{4} (x) - 2 sec^{2} (x) + 3 = 0

Rewrite using trig identities

3 + tan^{4} (x) - 2 sec^{2} (x)

Use the Pythagorean identity:

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

= 3 + tan^{4} (x) - 2 (tan^{2} (x) + 1)

Simplify

3 + tan^{4} (x) - 2 (tan^{2} (x) + 1) : tan^{4} (x) - 2 tan^{2} (x) + 1

3 + tan^{4} (x) - 2 (tan^{2} (x) + 1)

Expand

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

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

Apply the distributive law:

a (b + c) = ab + a c

a = - 2, b = tan^{2} (x), c = 1

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

Apply minus-plus rules

+ (- a) = - a

= - 2 tan^{2} (x) - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

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

= 3 + tan^{4} (x) - 2 tan^{2} (x) - 2

Simplify

3 + tan^{4} (x) - 2 tan^{2} (x) - 2 : tan^{4} (x) - 2 tan^{2} (x) + 1

3 + tan^{4} (x) - 2 tan^{2} (x) - 2

Group like terms

= tan^{4} (x) - 2 tan^{2} (x) + 3 - 2

Add/Subtract the numbers:

3 - 2 = 1

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

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

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

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

Solve by substitution

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

Let:

tan (x) = u

1 + u^{4} - 2 u^{2} = 0

1 + u^{4} - 2 u^{2} = 0 : u = 1, u = - 1

1 + u^{4} - 2 u^{2} = 0

Write in the standard form

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

u^{4} - 2 u^{2} + 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

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

Solve

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

v_{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

Apply exponent rule:

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

n

is even

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

= 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

v_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

v = \frac{- ( - 2 )}{2 \cdot 1}

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

v = 1

The solution to the quadratic equation is:

v = 1

v = 1

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = 1, u = - 1

Substitute back

u = tan (x)

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

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

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) = - 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{π}{4} + πn, x = \frac{3 π}{4} + πn

Popular Examples

sqrt(sin(x))=2sin(x)-1

sin (x) = 2 sin (x) - 1

tan(x)=-5/3

tan (x) = - \frac{5}{3}

3sin(θ)+4cos(θ)=3

3 sin (θ) + 4 cos (θ) = 3

cos(x)=(-5)/(13)

cos (x) = \frac{- 5}{13}

1=3cos(2θ)

1 = 3 cos (2 θ)

Frequently Asked Questions (FAQ)

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