What is the general solution for sec(2x)=tan(2x)+1 ?

The general solution for sec(2x)=tan(2x)+1 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 >

sec(2x)=tan(2x)+1

Solution

sec (2 x) = tan (2 x) + 1

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

sec (2 x) = tan (2 x) + 1

Subtract

tan (2 x) + 1

from both sides

sec (2 x) - tan (2 x) - 1 = 0

Express with sin, cos

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

Simplify

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

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

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 )}

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (2 x) = cos (2 x)

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

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

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

1 - sin (2 x) - cos (2 x) = 0

Add

cos (2 x)

to both sides

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

Square both sides

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

Subtract

cos^{2} (2 x)

from both sides

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

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

Refine

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

(1 - sin (2 x) + cos (2 x)) (1 - sin (2 x) - cos (2 x)) = 0

Solving each part separately

1 - sin (2 x) + cos (2 x) = 0 or 1 - sin (2 x) - cos (2 x) = 0

1 - sin (2 x) + cos (2 x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn

1 - sin (2 x) + cos (2 x) = 0

Rewrite using trig identities

1 + cos (2 x) - sin (2 x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Use the Double Angle identity:

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

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

Simplify

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

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

Group like terms

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

1 - 1 = 0

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

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

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

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

2 cos (x)

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

2 cos (x) (cos (x) - sin (x)) = 0

Solving each part separately

cos (x) = 0 or cos (x) - sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

cos (x) - sin (x) = 0

Rewrite using trig identities

cos (x) - sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

1 - tan (x) = 0

1 - tan (x) = 0

Move

1

to the right side

1 - tan (x) = 0

Subtract

1

from both sides

1 - tan (x) - 1 = 0 - 1

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (x) = 1

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

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn

1 - sin (2 x) - cos (2 x) = 0 : x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

1 - sin (2 x) - cos (2 x) = 0

Rewrite using trig identities

1 - cos (2 x) - sin (2 x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

1 - 1 = 0

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

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

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

Factor

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

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

Apply exponent rule:

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

sin^{2} (x) = sin (x) sin (x)

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

Factor out common term

2 sin (x)

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

2 sin (x) (sin (x) - cos (x)) = 0

Solving each part separately

sin (x) = 0 or sin (x) - cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

sin (x) - cos (x) = 0 : x = \frac{π}{4} + πn

sin (x) - cos (x) = 0

Rewrite using trig identities

sin (x) - cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

tan (x) - 1 = 0

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to both sides

tan (x) - 1 + 1 = 0 + 1

Simplify

tan (x) = 1

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

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn, x = 2 πn, x = π + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sec (2 x) = tan (2 x) + 1

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

False

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

sec (2 x) = tan (2 x) + 1

plug in

x = \frac{π}{2} + 2 π 1

sec (2 (\frac{π}{2} + 2 π 1)) = tan (2 (\frac{π}{2} + 2 π 1)) + 1

Refine

- 1 = 1

\Rightarrow False

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

sec (2 x) = tan (2 x) + 1

plug in

x = \frac{3 π}{2} + 2 π 1

sec (2 (\frac{3 π}{2} + 2 π 1)) = tan (2 (\frac{3 π}{2} + 2 π 1)) + 1

Refine

- 1 = 1

\Rightarrow False

Check the solution

\frac{π}{4} + πn :

True

\frac{π}{4} + πn

Plug in

n = 1

\frac{π}{4} + π 1

For

sec (2 x) = tan (2 x) + 1

plug in

x = \frac{π}{4} + π 1

sec (2 (\frac{π}{4} + π 1)) = tan (2 (\frac{π}{4} + π 1)) + 1

Refine

\infty = \infty

\Rightarrow True

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

sec (2 x) = tan (2 x) + 1

plug in

x = 2 π 1

sec (2 \cdot 2 π 1) = tan (2 \cdot 2 π 1) + 1

Refine

1 = 1

\Rightarrow True

Check the solution

π + 2 πn :

True

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

sec (2 x) = tan (2 x) + 1

plug in

x = π + 2 π 1

sec (2 (π + 2 π 1)) = tan (2 (π + 2 π 1)) + 1

Refine

1 = 1

\Rightarrow True

x = \frac{π}{4} + πn, x = 2 πn, x = π + 2 πn

Since the equation is undefined for:

\frac{π}{4} + πn

x = 2 πn, x = π + 2 πn

Popular Examples

sec^2(x)+3sec(x)+2=0

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

3sin(x)cos(x)-2cos(x)=0

3 sin (x) cos (x) - 2 cos (x) = 0

sin(x)=cos(x)+1

sin (x) = cos (x) + 1

5sin(x)+2=0

5 sin (x) + 2 = 0

4-sec^2(x)=0

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

Frequently Asked Questions (FAQ)

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