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

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

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sec(x)tan(x)= 2/3

Solution

sec (x) tan (x) = \frac{2}{3}

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

sec (x) tan (x) = \frac{2}{3}

Subtract

\frac{2}{3}

from both sides

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

Simplify

sec (x) tan (x) - \frac{2}{3} : \frac{3 sec ( x ) tan ( x ) - 2}{3}

sec (x) tan (x) - \frac{2}{3}

Convert element to fraction:

sec (x) tan (x) = \frac{sec ( x ) tan ( x ) 3}{3}

= \frac{sec ( x ) tan ( x ) \cdot 3}{3} - \frac{2}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{sec ( x ) tan ( x ) \cdot 3 - 2}{3}

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

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

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

Express with sin, cos

- 2 + 3 sec (x) tan (x)

Use the basic trigonometric identity:

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

= - 2 + 3 \cdot \frac{1}{cos ( x )} tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Convert element to fraction:

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

- (1 - u^{2}) \cdot 2 + 3 u = 0

- (1 - u^{2}) \cdot 2 + 3 u = 0 : u = \frac{1}{2}, u = - 2

- (1 - u^{2}) \cdot 2 + 3 u = 0

Expand

- (1 - u^{2}) \cdot 2 + 3 u : - 2 + 2 u^{2} + 3 u

- (1 - u^{2}) \cdot 2 + 3 u

= - 2 (1 - u^{2}) + 3 u

Expand

- 2 (1 - u^{2}) : - 2 + 2 u^{2}

- 2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

= - 2 \cdot 1 - (- 2) u^{2}

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 5}{2 \cdot 2}, u_{2} = \frac{- 3 - 5}{2 \cdot 2}

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 5 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 3 - 5}{2 \cdot 2} : - 2

\frac{- 3 - 5}{2 \cdot 2}

Subtract the numbers:

- 3 - 5 = - 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

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

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csc^{2} (x) - csc (x) + 9 = 11

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Frequently Asked Questions (FAQ)

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