What is the general solution for (2sec^2(x)-1)/(sec^2(x))=sec^2(x) ?

The general solution for (2sec^2(x)-1)/(sec^2(x))=sec^2(x) is x=2pin,x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

(2sec^2(x)-1)/(sec^2(x))=sec^2(x)

Solution

\frac{2 sec ^{2} ( x ) - 1}{sec ^{2} ( x )} = sec^{2} (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

\frac{2 sec ^{2} ( x ) - 1}{sec ^{2} ( x )} = sec^{2} (x)

Solve by substitution

\frac{2 sec ^{2} ( x ) - 1}{sec ^{2} ( x )} = sec^{2} (x)

Let:

sec (x) = u

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

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

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

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

Simplify

u^{2} u^{2} : u^{4}

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

Apply exponent rule:

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

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

= u^{2 + 2}

Add the numbers:

2 + 2 = 4

= u^{4}

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

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

Solve

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

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

Switch sides

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

Move

1

to the left side

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

Add

1

to both sides

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

Simplify

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

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

Move

2 u^{2}

to the left side

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

Subtract

2 u^{2}

from both sides

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

Simplify

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

u^{4} + 1 - 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

u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1, u = - 1

Substitute back

u = sec (x)

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

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

sec (x) = 1 : x = 2 πn

sec (x) = 1

General solutions for

sec (x) = 1

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 1

General solutions for

sec (x) = - 1

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Graph

Popular Examples

9(sin(x)-0.6pi)+8=0

solvefor t,v=100cos(50t+20)v

cos(2x)=sin(pi/2-x)

sin(x)=sin(50)

solvefor n,y=-sin(2((3pi)/4+pin))2

Frequently Asked Questions (FAQ)

What is the general solution for (2sec^2(x)-1)/(sec^2(x))=sec^2(x) ?
The general solution for (2sec^2(x)-1)/(sec^2(x))=sec^2(x) is x=2pin,x=pi+2pin