What is the general solution for tan(x)+sec(x)=sqrt(3) ?

The general solution for tan(x)+sec(x)=sqrt(3) is x= pi/6+2pin

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Popular Trigonometry >

tan(x)+sec(x)=sqrt(3)

Solution

tan (x) + sec (x) = 3

Solution

x = \frac{π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (x) + sec (x) = 3

Subtract

3

from both sides

tan (x) + sec (x) - 3 = 0

Express with sin, cos

\frac{sin ( x )}{cos ( x )} + \frac{1}{cos ( x )} - 3 = 0

Simplify

\frac{sin ( x )}{cos ( x )} + \frac{1}{cos ( x )} - 3 : \frac{sin ( x ) + 1 - 3 cos ( x )}{cos ( x )}

\frac{sin ( x )}{cos ( x )} + \frac{1}{cos ( x )} - 3

Combine the fractions

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

Apply rule

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

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

= \frac{sin ( x ) + 1}{cos ( x )} - 3

Convert element to fraction:

3 = \frac{3 cos ( x )}{cos ( x )}

= \frac{sin ( x ) + 1}{cos ( x )} - \frac{3 cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( x ) + 1 - 3 cos ( x )}{cos ( x )}

\frac{sin ( x ) + 1 - 3 cos ( x )}{cos ( x )} = 0

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

sin (x) + 1 - 3 cos (x) = 0

Add

3 cos (x)

to both sides

sin (x) + 1 = 3 cos (x)

Square both sides

(sin (x) + 1)^{2} = (3 cos (x))^{2}

Subtract

(3 cos (x))^{2}

from both sides

(sin (x) + 1)^{2} - 3 cos^{2} (x) = 0

Rewrite using trig identities

(1 + sin (x))^{2} - 3 cos^{2} (x)

Use the Pythagorean identity:

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

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

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

Simplify

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

(1 + sin (x))^{2} - 3 (1 - sin^{2} (x))

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = sin (x)

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

Simplify

1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x) : 1 + 2 sin (x) + sin^{2} (x)

1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

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

Expand

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

- 3 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (x)

= - 3 \cdot 1 - (- 3) sin^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (x)

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

Simplify

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

1 + 2 sin (x) + sin^{2} (x) - 3 + 3 sin^{2} (x)

Group like terms

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

Add similar elements:

sin^{2} (x) + 3 sin^{2} (x) = 4 sin^{2} (x)

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

Add/Subtract the numbers:

1 - 3 = - 2

= 4 sin^{2} (x) + 2 sin (x) - 2

= 4 sin^{2} (x) + 2 sin (x) - 2

= 4 sin^{2} (x) + 2 sin (x) - 2

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = 2, c = - 2

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

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

2^{2} - 4 \cdot 4 (- 2) = 6

2^{2} - 4 \cdot 4 (- 2)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 2 + 6}{2 \cdot 4}, u_{2} = \frac{- 2 - 6}{2 \cdot 4}

u = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

\frac{- 2 + 6}{2 \cdot 4}

Add/Subtract the numbers:

- 2 + 6 = 4

= \frac{4}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{- 2 - 6}{2 \cdot 4} : - 1

\frac{- 2 - 6}{2 \cdot 4}

Subtract the numbers:

- 2 - 6 = - 8

= \frac{- 8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{8}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

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 = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

sin (x) = - 1

General solutions for

sin (x) = - 1

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 = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan (x) + sec (x) = 3

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

Check the solution

\frac{π}{6} + 2 πn :

True

\frac{π}{6} + 2 πn

Plug in

n = 1

\frac{π}{6} + 2 π 1

For

tan (x) + sec (x) = 3

plug in

x = \frac{π}{6} + 2 π 1

tan (\frac{π}{6} + 2 π 1) + sec (\frac{π}{6} + 2 π 1) = 3

Refine

1.73205 \dots = 1.73205 \dots

\Rightarrow True

Check the solution

\frac{5 π}{6} + 2 πn :

False

\frac{5 π}{6} + 2 πn

Plug in

n = 1

\frac{5 π}{6} + 2 π 1

For

tan (x) + sec (x) = 3

plug in

x = \frac{5 π}{6} + 2 π 1

tan (\frac{5 π}{6} + 2 π 1) + sec (\frac{5 π}{6} + 2 π 1) = 3

Refine

- 1.73205 \dots = 1.73205 \dots

\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

tan (x) + sec (x) = 3

plug in

x = \frac{3 π}{2} + 2 π 1

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

Undefined

\Rightarrow False

x = \frac{π}{6} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x)+sec(x)=sqrt(3) ?
The general solution for tan(x)+sec(x)=sqrt(3) is x= pi/6+2pin