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

The general solution for cos(2x)=sec(x) is x=2pin

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Popular Trigonometry >

cos(2x)=sec(x)

Solution

cos (2 x) = sec (x)

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) = sec (x)

Subtract

sec (x)

from both sides

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

Rewrite using trig identities

cos (2 x) - sec (x)

Use the basic trigonometric identity:

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

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

Use the Double Angle identity:

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

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

- 1 - \frac{1}{cos ( x )} + 2 cos^{2} (x) = 0

Solve by substitution

- 1 - \frac{1}{cos ( x )} + 2 cos^{2} (x) = 0

Let:

cos (x) = u

- 1 - \frac{1}{u} + 2 u^{2} = 0

- 1 - \frac{1}{u} + 2 u^{2} = 0 : u = 1, u = - \frac{1}{2} + i \frac{1}{2}, u = - \frac{1}{2} - i \frac{1}{2}

- 1 - \frac{1}{u} + 2 u^{2} = 0

Multiply both sides by

u

- 1 - \frac{1}{u} + 2 u^{2} = 0

Multiply both sides by

u

- 1 \cdot u - \frac{1}{u} u + 2 u^{2} u = 0 \cdot u

Simplify

- 1 \cdot u - \frac{1}{u} u + 2 u^{2} u = 0 \cdot u

Simplify

- 1 \cdot u : - u

- 1 \cdot u

Multiply:

1 \cdot u = u

= - 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

2 u^{2} u : 2 u^{3}

2 u^{2} u

Apply exponent rule:

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

u^{2} u = u^{2 + 1}

= 2 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 2 u^{3}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

- u - 1 + 2 u^{3} = 0

- u - 1 + 2 u^{3} = 0

- u - 1 + 2 u^{3} = 0

Solve

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

- u - 1 + 2 u^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{3} - u - 1 = 0

Factor

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

2 u^{3} - u - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 2

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2}

\frac{1}{1}

is a root of the expression, so factor out

u - 1

= (u - 1) \frac{2 u ^{3} - u - 1}{u - 1}

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

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

Divide

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

Divide the leading coefficients of the numerator

2 u^{3} - u - 1

and the divisor

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

Quotient = 2 u^{2}

Multiply

u - 1

2 u^{2} : 2 u^{3} - 2 u^{2}

Subtract

2 u^{3} - 2 u^{2}

from

2 u^{3} - u - 1

to get new remainder

Remainder = 2 u^{2} - u - 1

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

2 u^{2} - u - 1

and the divisor

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

Quotient = 2 u

Multiply

u - 1

2 u : 2 u^{2} - 2 u

Subtract

2 u^{2} - 2 u

from

2 u^{2} - u - 1

to get new remainder

Remainder = u - 1

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

u - 1

and the divisor

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

Quotient = 1

Multiply

u - 1

1 : u - 1

Subtract

u - 1

from

u - 1

to get new remainder

Remainder = 0

Therefore

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

= 2 u^{2} + 2 u + 1

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

Solve

2 u^{2} + 2 u + 1 = 0 : u = - \frac{1}{2} + i \frac{1}{2}, u = - \frac{1}{2} - i \frac{1}{2}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 2, c = 1

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

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

Simplify

2^{2} - 4 \cdot 2 \cdot 1 : 2 i

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} - 8

Apply imaginary number rule:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

Add/Subtract the numbers:

- 4 + 8 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= 2 i

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

Separate the solutions

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

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

\frac{- 2 + 2 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 i}{4}

Factor

- 2 + 2 i : 2 (- 1 + i)

- 2 + 2 i

Rewrite as

= - 2 \cdot 1 + 2 i

Factor out common term

2

= 2 (- 1 + i)

= \frac{2 ( - 1 + i )}{4}

Cancel the common factor:

2

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

Rewrite

\frac{- 1 + i}{2}

in standard complex form:

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

\frac{- 1 + i}{2}

Apply the fraction rule:

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

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

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

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

u = \frac{- 2 - 2 i}{2 \cdot 2} : - \frac{1}{2} - i \frac{1}{2}

\frac{- 2 - 2 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 - 2 i}{4}

Factor

- 2 - 2 i : - 2 (1 + i)

- 2 - 2 i

Rewrite as

= - 2 \cdot 1 - 2 i

Factor out common term

2

= - 2 (1 + i)

= - \frac{2 ( 1 + i )}{4}

Cancel the common factor:

2

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

Rewrite

- \frac{1 + i}{2}

in standard complex form:

- \frac{1}{2} - \frac{1}{2} i

- \frac{1 + i}{2}

Apply the fraction rule:

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

\frac{1 + i}{2} = - (\frac{1}{2}) - (\frac{i}{2})

= - (\frac{1}{2}) - (\frac{i}{2})

Remove parentheses:

(a) = a

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

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

The solutions to the quadratic equation are:

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

The solutions are

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 1 - \frac{1}{u} + 2 u^{2}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = cos (x)

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

No Solution

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

No Solution

cos (x) = - \frac{1}{2} - i \frac{1}{2} :

No Solution

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

No Solution

Combine all the solutions

x = 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)=sec(x) ?
The general solution for cos(2x)=sec(x) is x=2pin