What is the general solution for cos(2x)+sec(2x)=11 ?

The general solution for cos(2x)+sec(2x)=11 is x=(1.47899…)/2+pin,x=pi-(1.47899…)/2+pin

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cos(2x)+sec(2x)=11

Solution

cos (2 x) + sec (2 x) = 11

Solution

x = \frac{1.47899 \dots}{2} + πn, x = π - \frac{1.47899 \dots}{2} + πn

Degrees

x = 42.37006 \dots^{\circ} + 18 0^{\circ} n, x = 137.62993 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 x) + sec (2 x) = 11

Subtract

11

from both sides

cos (2 x) + sec (2 x) - 11 = 0

Rewrite using trig identities

- 11 + cos (2 x) + sec (2 x)

Use the basic trigonometric identity:

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

= - 11 + \frac{1}{sec ( 2 x )} + sec (2 x)

- 11 + \frac{1}{sec ( 2 x )} + sec (2 x) = 0

Solve by substitution

- 11 + \frac{1}{sec ( 2 x )} + sec (2 x) = 0

Let:

sec (2 x) = u

- 11 + \frac{1}{u} + u = 0

- 11 + \frac{1}{u} + u = 0 : u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

- 11 + \frac{1}{u} + u = 0

Multiply both sides by

u

- 11 + \frac{1}{u} + u = 0

Multiply both sides by

u

- 11 u + \frac{1}{u} u + uu = 0 \cdot u

Simplify

- 11 u + \frac{1}{u} u + uu = 0 \cdot u

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

uu : u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

- 11 u + 1 + u^{2} = 0

- 11 u + 1 + u^{2} = 0

- 11 u + 1 + u^{2} = 0

Solve

- 11 u + 1 + u^{2} = 0 : u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

- 11 u + 1 + u^{2} = 0

Write in the standard form

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

u^{2} - 11 u + 1 = 0

Solve with the quadratic formula

u^{2} - 11 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 11, c = 1

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 1 = 313

(- 11)^{2} - 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 11)^{2} = 1 1^{2}

= 1 1^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 1^{2} - 4

1 1^{2} = 121

= 121 - 4

Subtract the numbers:

121 - 4 = 117

= 117

Prime factorization of

117 : 3^{2} \cdot 13

117

117

divides by

3117 = 39 \cdot 3

= 3 \cdot 39

39

divides by

339 = 13 \cdot 3

= 3 \cdot 3 \cdot 13

3, 13

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 3 \cdot 13

= 3^{2} \cdot 13

= 3^{2} \cdot 13

Apply radical rule:

= 13 3^{2}

Apply radical rule:

3^{2} = 3

= 313

u_{1, 2} = \frac{- ( - 11 ) \pm 3 13}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 3 13}{2 \cdot 1}, u_{2} = \frac{- ( - 11 ) - 3 13}{2 \cdot 1}

u = \frac{- ( - 11 ) + 3 13}{2 \cdot 1} : \frac{11 + 3 13}{2}

\frac{- ( - 11 ) + 3 13}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{11 + 3 13}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{11 + 3 13}{2}

u = \frac{- ( - 11 ) - 3 13}{2 \cdot 1} : \frac{11 - 3 13}{2}

\frac{- ( - 11 ) - 3 13}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{11 - 3 13}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{11 - 3 13}{2}

The solutions to the quadratic equation are:

u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 11 + \frac{1}{u} + u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

Substitute back

u = sec (2 x)

sec (2 x) = \frac{11 + 3 13}{2}, sec (2 x) = \frac{11 - 3 13}{2}

sec (2 x) = \frac{11 + 3 13}{2}, sec (2 x) = \frac{11 - 3 13}{2}

sec (2 x) = \frac{11 + 3 13}{2} : x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn, x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

sec (2 x) = \frac{11 + 3 13}{2}

Apply trig inverse properties

sec (2 x) = \frac{11 + 3 13}{2}

General solutions for

sec (2 x) = \frac{11 + 3 13}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn, 2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn, 2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn

Solve

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn : x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn

Divide both sides by

2

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

Solve

2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn : x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn

Divide both sides by

2

2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + \frac{2 πn}{2}

Simplify

x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn, x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

sec (2 x) = \frac{11 - 3 13}{2} :

No Solution

sec (2 x) = \frac{11 - 3 13}{2}

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

Combine all the solutions

x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn, x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

Show solutions in decimal form

x = \frac{1.47899 \dots}{2} + πn, x = π - \frac{1.47899 \dots}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)+sec(2x)=11 ?
The general solution for cos(2x)+sec(2x)=11 is x=(1.47899…)/2+pin,x=pi-(1.47899…)/2+pin