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

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

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

Solution

sec (x) tan (x) = 23

Solution

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n

Solution steps

sec (x) tan (x) = 23

Subtract

23

from both sides

sec (x) tan (x) - 23 = 0

Express with sin, cos

- 23 + sec (x) tan (x)

Use the basic trigonometric identity:

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

= - 23 + \frac{1}{cos ( x )} tan (x)

Use the basic trigonometric identity:

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

= - 23 + \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Simplify

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

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

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

= - 23 + \frac{sin ( x )}{cos ^{2} ( x )}

Convert element to fraction:

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

sin (x) - (1 - sin^{2} (x)) \cdot 23 = 0

Solve by substitution

sin (x) - (1 - sin^{2} (x)) \cdot 23 = 0

Let:

sin (x) = u

u - (1 - u^{2}) \cdot 23 = 0

u - (1 - u^{2}) \cdot 23 = 0 : u = \frac{3}{2}, u = - \frac{2 3}{3}

u - (1 - u^{2}) \cdot 23 = 0

Expand

u - (1 - u^{2}) \cdot 23 : u - 23 + 23 u^{2}

u - (1 - u^{2}) \cdot 23

= u - 23 (1 - u^{2})

Expand

- 23 (1 - u^{2}) : - 23 + 23 u^{2}

- 23 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 23, b = 1, c = u^{2}

= - 23 \cdot 1 - (- 23) u^{2}

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 \cdot 3 + 23 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= - 23 + 23 u^{2}

= u - 23 + 23 u^{2}

u - 23 + 23 u^{2} = 0

Write in the standard form

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

23 u^{2} + u - 23 = 0

Solve with the quadratic formula

23 u^{2} + u - 23 = 0

Quadratic Equation Formula:

For

a = 23, b = 1, c = - 23

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

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

1^{2} - 4 \cdot 23 (- 23) = 7

1^{2} - 4 \cdot 23 (- 23)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 23 (- 23)

Apply rule

- (- a) = a

= 1 + 4 \cdot 23 \cdot 23

4 \cdot 23 \cdot 23 = 48

4 \cdot 23 \cdot 23

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 1633

Apply radical rule:

a a = a

33 = 3

= 16 \cdot 3

Multiply the numbers:

16 \cdot 3 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 2 3}

Separate the solutions

u_{1} = \frac{- 1 + 7}{2 \cdot 2 3}, u_{2} = \frac{- 1 - 7}{2 \cdot 2 3}

u = \frac{- 1 + 7}{2 \cdot 2 3} : \frac{3}{2}

\frac{- 1 + 7}{2 \cdot 2 3}

Add/Subtract the numbers:

- 1 + 7 = 6

= \frac{6}{2 \cdot 2 3}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{4 3}

Cancel the common factor:

2

= \frac{3}{2 3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{3}{2 \cdot 3 ^{\frac{1}{2}}}

Apply exponent rule:

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

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= \frac{3 ^{1 - \frac{1}{2}}}{2}

Subtract the numbers:

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

= \frac{3 ^{\frac{1}{2}}}{2}

Apply radical rule:

3^{\frac{1}{2}} = 3

= \frac{3}{2}

u = \frac{- 1 - 7}{2 \cdot 2 3} : - \frac{2 3}{3}

\frac{- 1 - 7}{2 \cdot 2 3}

Subtract the numbers:

- 1 - 7 = - 8

= \frac{- 8}{2 \cdot 2 3}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4 3}

Apply the fraction rule:

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

= - \frac{8}{4 3}

Divide the numbers:

\frac{8}{4} = 2

= - \frac{2}{3}

Rationalize

- \frac{2}{3} : - \frac{2 3}{3}

- \frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{2 3}{3}

= - \frac{2 3}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

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

General solutions for

sin (x) = \frac{3}{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}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

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

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

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cos(x+60)=-sin(x)

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cos^5(x)+cos(x)=2cos^2(x)

cos^{5} (x) + cos (x) = 2 cos^{2} (x)

Frequently Asked Questions (FAQ)

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