What is the general solution for (tan(x)-sec(x))^2=3 ?

The general solution for (tan(x)-sec(x))^2=3 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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Popular Trigonometry >

(tan(x)-sec(x))^2=3

Solution

(tan (x) - sec (x))^{2} = 3

Solution

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

(tan (x) - sec (x))^{2} = 3

Subtract

3

from both sides

(tan (x) - sec (x))^{2} - 3 = 0

Express with sin, cos

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

Simplify

(\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )})^{2} - 3 : \frac{( sin ( x ) - 1 ) ^{2} - 3 cos ^{2} ( x )}{cos ^{2} ( x )}

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

Apply exponent rule:

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

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

Convert element to fraction:

3 = \frac{3 cos ^{2} ( x )}{cos ^{2} ( x )}

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

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

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

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

Add

3 cos^{2} (x)

to both sides

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

Square both sides

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

Subtract

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

from both sides

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x) = 0

Factor

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x) : (sin^{2} (x) - 2 sin (x) + 1 + 3 cos^{2} (x)) (sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x))

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

Rewrite

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

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

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

Rewrite

9

3^{2}

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

Apply exponent rule:

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

cos^{4} (x) = (cos^{2} (x))^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

3^{2} (cos^{2} (x))^{2} = (3 cos^{2} (x))^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

Refine

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

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

Solving each part separately

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Multiply the numbers:

3 \cdot 1 = 3

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add the numbers:

1 + 3 = 4

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 4 = 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:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 6}{2 ( - 2 )}, u_{2} = \frac{- ( - 2 ) - 6}{2 ( - 2 )}

u = \frac{- ( - 2 ) + 6}{2 ( - 2 )} : - 2

\frac{- ( - 2 ) + 6}{2 ( - 2 )}

Remove parentheses:

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

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

Add the numbers:

2 + 6 = 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{- 4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 2 ) - 6}{2 ( - 2 )} : 1

\frac{- ( - 2 ) - 6}{2 ( - 2 )}

Remove parentheses:

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

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

Subtract the numbers:

2 - 6 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - 2, u = 1

Substitute back

u = sin (x)

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

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

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 1 : x = \frac{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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn

sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

= 4 sin^{2} (x) - 2 sin (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 = 1, u = - \frac{1}{2}

- 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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 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:

n a^{n} = a

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} : 1

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 6 = 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 6 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = 1 : x = \frac{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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

(tan (x) - sec (x))^{2} = 3

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

(tan (x) - sec (x))^{2} = 3

plug in

x = \frac{π}{2} + 2 π 1

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

Undefined

\Rightarrow False

Check the solution

\frac{7 π}{6} + 2 πn :

True

\frac{7 π}{6} + 2 πn

Plug in

n = 1

\frac{7 π}{6} + 2 π 1

For

(tan (x) - sec (x))^{2} = 3

plug in

x = \frac{7 π}{6} + 2 π 1

(tan (\frac{7 π}{6} + 2 π 1) - sec (\frac{7 π}{6} + 2 π 1))^{2} = 3

Refine

3 = 3

\Rightarrow True

Check the solution

\frac{11 π}{6} + 2 πn :

True

\frac{11 π}{6} + 2 πn

Plug in

n = 1

\frac{11 π}{6} + 2 π 1

For

(tan (x) - sec (x))^{2} = 3

plug in

x = \frac{11 π}{6} + 2 π 1

(tan (\frac{11 π}{6} + 2 π 1) - sec (\frac{11 π}{6} + 2 π 1))^{2} = 3

Refine

3 = 3

\Rightarrow True

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (tan(x)-sec(x))^2=3 ?
The general solution for (tan(x)-sec(x))^2=3 is x=(7pi)/6+2pin,x=(11pi)/6+2pin