What is the general solution for tan(x)+3cot(x)=5sec(x) ?

The general solution for tan(x)+3cot(x)=5sec(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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Popular Trigonometry >

tan(x)+3cot(x)=5sec(x)

Solution

tan (x) + 3 cot (x) = 5 sec (x)

Solution

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

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (x) + 3 cot (x) = 5 sec (x)

Subtract

5 sec (x)

from both sides

tan (x) + 3 cot (x) - 5 sec (x) = 0

Express with sin, cos

tan (x) + 3 cot (x) - 5 sec (x)

Use the basic trigonometric identity:

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

= \frac{sin ( x )}{cos ( x )} + 3 cot (x) - 5 sec (x)

Use the basic trigonometric identity:

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

= \frac{sin ( x )}{cos ( x )} + 3 \cdot \frac{cos ( x )}{sin ( x )} - 5 sec (x)

Use the basic trigonometric identity:

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

= \frac{sin ( x )}{cos ( x )} + 3 \cdot \frac{cos ( x )}{sin ( x )} - 5 \cdot \frac{1}{cos ( x )}

Simplify

\frac{sin ( x )}{cos ( x )} + 3 \cdot \frac{cos ( x )}{sin ( x )} - 5 \cdot \frac{1}{cos ( x )} : \frac{sin ( x ) ( sin ( x ) - 5 ) + 3 cos ^{2} ( x )}{cos ( x ) sin ( x )}

\frac{sin ( x )}{cos ( x )} + 3 \cdot \frac{cos ( x )}{sin ( x )} - 5 \cdot \frac{1}{cos ( x )}

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

3 \cdot \frac{cos ( x )}{sin ( x )}

Multiply fractions:

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

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

5 \cdot \frac{1}{cos ( x )} = \frac{5}{cos ( x )}

5 \cdot \frac{1}{cos ( x )}

Multiply fractions:

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

= \frac{1 \cdot 5}{cos ( x )}

Multiply the numbers:

1 \cdot 5 = 5

= \frac{5}{cos ( x )}

= \frac{sin ( x )}{cos ( x )} + \frac{3 cos ( x )}{sin ( x )} - \frac{5}{cos ( x )}

Combine the fractions

\frac{sin ( x )}{cos ( x )} - \frac{5}{cos ( x )} : \frac{sin ( x ) - 5}{cos ( x )}

Apply rule

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

= \frac{sin ( x ) - 5}{cos ( x )}

= \frac{sin ( x ) - 5}{cos ( x )} + \frac{3 cos ( x )}{sin ( x )}

Least Common Multiplier of

cos (x), sin (x) : cos (x) sin (x)

cos (x), sin (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

sin (x)

= cos (x) sin (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (x) sin (x)

For

\frac{sin ( x ) - 5}{cos ( x )} :

multiply the denominator and numerator by

sin (x)

\frac{sin ( x ) - 5}{cos ( x )} = \frac{( sin ( x ) - 5 ) sin ( x )}{cos ( x ) sin ( x )}

For

\frac{cos ( x ) \cdot 3}{sin ( x )} :

multiply the denominator and numerator by

cos (x)

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

= \frac{( sin ( x ) - 5 ) sin ( x )}{cos ( x ) sin ( x )} + \frac{3 cos ^{2} ( x )}{cos ( x ) sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{( sin ( x ) - 5 ) sin ( x ) + 3 cos ^{2} ( x )}{cos ( x ) sin ( x )}

= \frac{sin ( x ) ( sin ( x ) - 5 ) + 3 cos ^{2} ( x )}{cos ( x ) sin ( x )}

\frac{( - 5 + sin ( x ) ) sin ( x ) + 3 cos ^{2} ( x )}{cos ( x ) sin ( x )} = 0

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= (- 5 + sin (x)) sin (x) + 3 (1 - sin^{2} (x))

Simplify

(- 5 + sin (x)) sin (x) + 3 (1 - sin^{2} (x)) : - 5 sin (x) - 2 sin^{2} (x) + 3

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

= sin (x) (- 5 + sin (x)) + 3 (1 - sin^{2} (x))

Expand

sin (x) (- 5 + sin (x)) : - 5 sin (x) + sin^{2} (x)

sin (x) (- 5 + sin (x))

Apply the distributive law:

a (b + c) = ab + a c

a = sin (x), b = - 5, c = sin (x)

= sin (x) (- 5) + sin (x) sin (x)

Apply minus-plus rules

+ (- a) = - a

= - 5 sin (x) + sin (x) sin (x)

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= - 5 sin (x) + sin^{2} (x)

= - 5 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)

Multiply the numbers:

3 \cdot 1 = 3

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

= - 5 sin (x) + sin^{2} (x) + 3 - 3 sin^{2} (x)

Simplify

- 5 sin (x) + sin^{2} (x) + 3 - 3 sin^{2} (x) : - 5 sin (x) - 2 sin^{2} (x) + 3

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

Group like terms

= - 5 sin (x) + sin^{2} (x) - 3 sin^{2} (x) + 3

Add similar elements:

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

= - 5 sin (x) - 2 sin^{2} (x) + 3

= - 5 sin (x) - 2 sin^{2} (x) + 3

= - 5 sin (x) - 2 sin^{2} (x) + 3

3 - 2 sin^{2} (x) - 5 sin (x) = 0

Solve by substitution

3 - 2 sin^{2} (x) - 5 sin (x) = 0

Let:

sin (x) = u

3 - 2 u^{2} - 5 u = 0

3 - 2 u^{2} - 5 u = 0 : u = - 3, u = \frac{1}{2}

3 - 2 u^{2} - 5 u = 0

Write in the standard form

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

- 2 u^{2} - 5 u + 3 = 0

Solve with the quadratic formula

- 2 u^{2} - 5 u + 3 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 5, c = 3

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

(- 5)^{2} - 4 (- 2) \cdot 3 = 7

(- 5)^{2} - 4 (- 2) \cdot 3

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 2 \cdot 3

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 7}{2 ( - 2 )}, u_{2} = \frac{- ( - 5 ) - 7}{2 ( - 2 )}

u = \frac{- ( - 5 ) + 7}{2 ( - 2 )} : - 3

\frac{- ( - 5 ) + 7}{2 ( - 2 )}

Remove parentheses:

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

= \frac{5 + 7}{- 2 \cdot 2}

Add the numbers:

5 + 7 = 12

= \frac{12}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{- 4}

Apply the fraction rule:

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

= - \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= - 3

u = \frac{- ( - 5 ) - 7}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 5 ) - 7}{2 ( - 2 )}

Remove parentheses:

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

= \frac{5 - 7}{- 2 \cdot 2}

Subtract the numbers:

5 - 7 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = - 3, u = \frac{1}{2}

Substitute back

u = sin (x)

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

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

sin (x) = - 3 :

No Solution

sin (x) = - 3

- 1 \leq sin (x) \leq 1

No Solution

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

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x)+3cot(x)=5sec(x) ?
The general solution for tan(x)+3cot(x)=5sec(x) is x= pi/6+2pin,x=(5pi)/6+2pin