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

The general solution for -2=-2sec(x)+tan^2(x) is x=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-2=-2sec(x)+tan^2(x)

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Switch sides

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

Subtract

- 2

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

2 - 1 = 1

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

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

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

Solve by substitution

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

Let:

sec (x) = u

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

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

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

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}

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

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = sec (x)

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

Combine all the solutions

x = 2 πn

Popular Examples

tan^2(y)+1=cos^2(y)

tan^{2} (y) + 1 = cos^{2} (y)

4sin(2x)+csc(2x)=4

4 sin (2 x) + csc (2 x) = 4

cot(θ)= 9/40

cot (θ) = \frac{9}{40}

sin(a)=-1/(sqrt(2))

sin (a) = - \frac{1}{2}

sqrt(5)cos(x)-2sin(x)=0.5

5 cos (x) - 2 sin (x) = 0.5

Frequently Asked Questions (FAQ)

What is the general solution for -2=-2sec(x)+tan^2(x) ?
The general solution for -2=-2sec(x)+tan^2(x) is x=2pin