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

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

The general solution for sec^2(x)-2=tan^2(x),0<= x<= 2pi is No Solution for x\in\mathbb{R}

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

Solution

sec^{2} (x) - 2 = tan^{2} (x), 0 \leq x \leq 2 π

Solution

No Solution for x \in R

Solution steps

sec^{2} (x) - 2 = tan^{2} (x), 0 \leq x \leq 2 π

Subtract

tan^{2} (x)

from both sides

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

Express with sin, cos

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 2 + (\frac{1}{cos ( x )})^{2} - (\frac{sin ( x )}{cos ( x )})^{2}

Simplify

- 2 + (\frac{1}{cos ( x )})^{2} - (\frac{sin ( x )}{cos ( x )})^{2} : \frac{- 2 cos ^{2} ( x ) + 1 - sin ^{2} ( x )}{cos ^{2} ( x )}

- 2 + (\frac{1}{cos ( x )})^{2} - (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{1}{cos ( x )})^{2} = \frac{1}{cos ^{2} ( x )}

(\frac{1}{cos ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= - 2 + \frac{1}{cos ^{2} ( x )} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

Combine the fractions

\frac{1}{cos ^{2} ( x )} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )} : \frac{1 - sin ^{2} ( x )}{cos ^{2} ( x )}

Apply rule

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

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

= - 2 + \frac{- sin ^{2} ( x ) + 1}{cos ^{2} ( x )}

Convert element to fraction:

2 = \frac{2 cos ^{2} ( x )}{cos ^{2} ( x )}

= - \frac{2 cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{1 - sin ^{2} ( x )}{cos ^{2} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 2 cos ^{2} ( x ) + 1 - sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{- 2 cos ^{2} ( x ) + 1 - sin ^{2} ( x )}{cos ^{2} ( x )}

\frac{1 - sin ^{2} ( x ) - 2 cos ^{2} ( x )}{cos ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - sin^{2} (x) - 2 cos^{2} (x) = 0

Rewrite using trig identities

1 - sin^{2} (x) - 2 cos^{2} (x)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= - 2 cos^{2} (x) + cos^{2} (x)

Simplify

= - cos^{2} (x)

- cos^{2} (x) = 0

Divide both sides by

- 1

- cos^{2} (x) = 0

Divide both sides by

- 1

- cos^{2} (x) = 0

Divide both sides by

- 1

\frac{- cos ^{2} ( x )}{- 1} = \frac{0}{- 1}

Simplify

cos^{2} (x) = 0

cos^{2} (x) = 0

Apply rule

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

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

Since the equation is undefined for:

\frac{π}{2}, \frac{3 π}{2}

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)-2=tan^2(x),0<= x<= 2pi ?
The general solution for sec^2(x)-2=tan^2(x),0<= x<= 2pi is No Solution for x\in\mathbb{R}