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

The general solution for sqrt(2)=sec(x)tan(x) is x= pi/4+2pin,x=(3pi)/4+2pin

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

Solution

2 = sec (x) tan (x)

Solution

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

2 = sec (x) tan (x)

Switch sides

sec (x) tan (x) = 2

Subtract

2

from both sides

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

Express with sin, cos

- 2 + sec (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

- 2 + \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 )}

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

u - (1 - u^{2}) 2 = 0

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

u - (1 - u^{2}) 2 = 0

Expand

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

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

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

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

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

Multiply:

1 \cdot 2 = 2

= - 2 + 2 u^{2}

= u - 2 + 2 u^{2}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 2

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

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

1^{2} - 42 (- 2) = 3

1^{2} - 42 (- 2)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 42 (- 2)

Apply rule

- (- a) = a

= 1 + 422

422 = 8

422

Apply radical rule:

a a = a

22 = 2

= 4 \cdot 2

Multiply the numbers:

4 \cdot 2 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= \frac{1}{2}

Rationalize

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

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

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

Subtract the numbers:

- 1 - 3 = - 4

= \frac{- 4}{2 2}

Apply the fraction rule:

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

= - \frac{4}{2 2}

Divide the numbers:

\frac{4}{2} = 2

= \frac{2}{2}

Apply radical rule:

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

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

Apply exponent rule:

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

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

General solutions for

sin (x) = \frac{2}{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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

Popular Examples

cos(x)= 9/13

cos (x) = \frac{9}{13}

cos^2(x)=((6k-9))/(10)

cos^{2} (x) = \frac{( 6 k - 9 )}{10}

cot(x)+csc(x)=-1

cot (x) + csc (x) = - 1

cos^2(x)=((3k-3))/8

cos^{2} (x) = \frac{( 3 k - 3 )}{8}

8sin^2(x)-10cos(x)-11=0

8 sin^{2} (x) - 10 cos (x) - 11 = 0

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)=sec(x)tan(x) ?
The general solution for sqrt(2)=sec(x)tan(x) is x= pi/4+2pin,x=(3pi)/4+2pin