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sec(4x)-sec(2x)=2

Solution

sec (4 x) - sec (2 x) = 2

Solution

x = \frac{π}{2} + πn, x = \frac{0.62831 \dots}{2} + πn, x = π - \frac{0.62831 \dots}{2} + πn, x = \frac{1.88495 \dots}{2} + πn, x = - \frac{1.88495 \dots}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n, x = 1 8^{\circ} + 18 0^{\circ} n, x = 16 2^{\circ} + 18 0^{\circ} n, x = 5 4^{\circ} + 18 0^{\circ} n, x = - 5 4^{\circ} + 18 0^{\circ} n

Solution steps

sec (4 x) - sec (2 x) = 2

Subtract

2

from both sides

sec (4 x) - sec (2 x) - 2 = 0

Rewrite using trig identities

- 2 - sec (2 x) + sec (4 x)

Use the basic trigonometric identity:

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

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

cos (4 x) = 2 cos^{2} (2 x) - 1

cos (4 x)

Rewrite as

= cos (2 \cdot 2 x)

Use the Double Angle identity:

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

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

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

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

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

Solve by substitution

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

Let:

cos (2 x) = u

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

- 2 + \frac{1}{- 1 + 2 u ^{2}} - \frac{1}{u} = 0 : u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

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

Multiply by LCM

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

Find Least Common Multiplier of

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

- 1 + 2 u^{2}, u

Lowest Common Multiplier (LCM)

Factor the expressions

Factor

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

- 1 + 2 u^{2}

Rewrite

2 u^{2} - 1

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

2 u^{2} - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (2 u + 1) (2 u - 1)

Compute an expression comprised of factors that appear either in

(2 u + 1) (2 u - 1)

u

= u (2 u + 1) (2 u - 1)

Multiply by LCM=

u (2 u + 1) (2 u - 1)

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot u = u

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

Factor

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

2 u^{2} - 1

Rewrite

2 u^{2} - 1

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

2 u^{2} - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (2 u + 1) (2 u - 1)

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

Cancel

\frac{u ( 2 u + 1 ) ( 2 u - 1 )}{( 2 u + 1 ) ( 2 u - 1 )} : u

\frac{u ( 2 u + 1 ) ( 2 u - 1 )}{( 2 u + 1 ) ( 2 u - 1 )}

Cancel the common factor:

2 u + 1

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

Cancel the common factor:

2 u - 1

= u

= u

Simplify

- \frac{1}{u} u (2 u + 1) (2 u - 1) : - (2 u + 1) (2 u - 1)

- \frac{1}{u} u (2 u + 1) (2 u - 1)

Multiply fractions:

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

= - \frac{1 \cdot u ( 2 u + 1 ) ( 2 u - 1 )}{u}

Cancel the common factor:

u

= - 1 \cdot (2 u + 1) (2 u - 1)

Multiply:

1 \cdot (2 u + 1) = (2 u + 1)

= - (2 u + 1) (2 u - 1)

Simplify

0 \cdot u (2 u + 1) (2 u - 1) : 0

0 \cdot u (2 u + 1) (2 u - 1)

Apply rule

0 \cdot a = 0

= 0

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0

Solve

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0 : u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0

Factor

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) : - (u + 1) (4 u^{2} - 2 u - 1)

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1)

Expand

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) : - 4 u^{3} + 3 u - 2 u^{2} + 1

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1)

Expand

- 2 u (2 u + 1) (2 u - 1) : - 4 u^{3} + 2 u

Expand

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

(2 u + 1) (2 u - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 2 u, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (2 u)^{2} - 1

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

(2 u)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

(2)^{2} : 2

Apply radical rule:

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

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

= 2 u^{2}

= 2 u^{2} - 1

= 2 u^{2} - 1

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

Expand

- 2 u (2 u^{2} - 1) : - 4 u^{3} + 2 u

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Simplify

- 2 \cdot 2 u^{2} u + 2 \cdot 1 \cdot u : - 4 u^{3} + 2 u

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

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

2 \cdot 2 u^{2} u

Multiply the numbers:

2 \cdot 2 = 4

= 4 u^{2} u

Apply exponent rule:

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

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

= 4 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 4 u^{3}

2 \cdot 1 \cdot u = 2 u

2 \cdot 1 \cdot u

Multiply the numbers:

2 \cdot 1 = 2

= 2 u

= - 4 u^{3} + 2 u

= - 4 u^{3} + 2 u

= - 4 u^{3} + 2 u

= - 4 u^{3} + 2 u + u - (2 u + 1) (2 u - 1)

Expand

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

Expand

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

(2 u + 1) (2 u - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 2 u, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (2 u)^{2} - 1

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

(2 u)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

(2)^{2} : 2

Apply radical rule:

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

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

= 2 u^{2}

= 2 u^{2} - 1

= 2 u^{2} - 1

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 2 u^{2} + 1

= - 4 u^{3} + 2 u + u - 2 u^{2} + 1

Add similar elements:

2 u + u = 3 u

= - 4 u^{3} + 3 u - 2 u^{2} + 1

= - 4 u^{3} + 3 u - 2 u^{2} + 1

Factor

- 4 u^{3} - 2 u^{2} + 3 u + 1 : - (u + 1) (4 u^{2} - 2 u - 1)

- 4 u^{3} - 2 u^{2} + 3 u + 1

Factor out common term

- 1

= - (4 u^{3} + 2 u^{2} - 3 u - 1)

Factor

4 u^{3} + 2 u^{2} - 3 u - 1 : (u + 1) (4 u^{2} - 2 u - 1)

4 u^{3} + 2 u^{2} - 3 u - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 4

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4}

- \frac{1}{1}

is a root of the expression, so factor out

u + 1

= (u + 1) \frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} - 2 u - 1

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

Divide

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} : \frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

Divide the leading coefficients of the numerator

4 u^{3} + 2 u^{2} - 3 u - 1

and the divisor

u + 1 : \frac{4 u ^{3}}{u} = 4 u^{2}

Quotient = 4 u^{2}

Multiply

u + 1

4 u^{2} : 4 u^{3} + 4 u^{2}

Subtract

4 u^{3} + 4 u^{2}

from

4 u^{3} + 2 u^{2} - 3 u - 1

to get new remainder

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

Therefore

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

= 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

Divide

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

Divide the leading coefficients of the numerator

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

to get new remainder

Remainder = - u - 1

Therefore

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

= 4 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

= 4 u^{2} - 2 u - 1

= 4 u^{2} - 2 u - 1

= (u + 1) (4 u^{2} - 2 u - 1)

= - (u + 1) (4 u^{2} - 2 u - 1)

= - (u + 1) (4 u^{2} - 2 u - 1)

- (u + 1) (4 u^{2} - 2 u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u + 1 = 0 or 4 u^{2} - 2 u - 1 = 0

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

4 u^{2} - 2 u - 1 = 0 : u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 4 (- 1) = 25

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

Add the numbers:

4 + 16 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Apply radical rule:

2^{2} = 2

= 25

u_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 4}

u = \frac{- ( - 2 ) + 2 5}{2 \cdot 4} : \frac{1 + 5}{4}

\frac{- ( - 2 ) + 2 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 + 2 5}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 + 2 5}{8}

Factor

2 + 25 : 2 (1 + 5)

2 + 25

Rewrite as

= 2 \cdot 1 + 25

Factor out common term

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{8}

Cancel the common factor:

2

= \frac{1 + 5}{4}

u = \frac{- ( - 2 ) - 2 5}{2 \cdot 4} : \frac{1 - 5}{4}

\frac{- ( - 2 ) - 2 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 - 2 5}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 - 2 5}{8}

Factor

2 - 25 : 2 (1 - 5)

2 - 25

Rewrite as

= 2 \cdot 1 - 25

Factor out common term

2

= 2 (1 - 5)

= \frac{2 ( 1 - 5 )}{8}

Cancel the common factor:

2

= \frac{1 - 5}{4}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

The solutions are

u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

Verify Solutions

Find undefined (singularity) points:

u = \frac{1}{2}, u = - \frac{1}{2}, u = 0

Take the denominator(s) of

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

and compare to zero

Solve

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

- 1 + 2 u^{2} = 0

Move

1

to the right side

- 1 + 2 u^{2} = 0

Add

1

to both sides

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

Simplify

2 u^{2} = 1

2 u^{2} = 1

Divide both sides by

2

2 u^{2} = 1

Divide both sides by

2

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

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

\frac{1}{2}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{2}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{2}

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

- \frac{1}{2}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{2}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{2}

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

u = 0

The following points are undefined

u = \frac{1}{2}, u = - \frac{1}{2}, u = 0

Combine undefined points with solutions:

u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

Substitute back

u = cos (2 x)

cos (2 x) = - 1, cos (2 x) = \frac{1 + 5}{4}, cos (2 x) = \frac{1 - 5}{4}

cos (2 x) = - 1, cos (2 x) = \frac{1 + 5}{4}, cos (2 x) = \frac{1 - 5}{4}

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

cos (2 x) = - 1

General solutions for

cos (2 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}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

cos (2 x) = \frac{1 + 5}{4} : x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn, x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

cos (2 x) = \frac{1 + 5}{4}

Apply trig inverse properties

cos (2 x) = \frac{1 + 5}{4}

General solutions for

cos (2 x) = \frac{1 + 5}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 x = arccos (\frac{1 + 5}{4}) + 2 πn, 2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

2 x = arccos (\frac{1 + 5}{4}) + 2 πn, 2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

Solve

2 x = arccos (\frac{1 + 5}{4}) + 2 πn : x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

2 x = arccos (\frac{1 + 5}{4}) + 2 πn

Divide both sides by

2

2 x = arccos (\frac{1 + 5}{4}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( \frac{1 + 5}{4} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn : x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

Divide both sides by

2

2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1 + 5}{4} )}{2} + \frac{2 πn}{2}

Simplify

x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn, x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

cos (2 x) = \frac{1 - 5}{4} : x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn, x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

cos (2 x) = \frac{1 - 5}{4}

Apply trig inverse properties

cos (2 x) = \frac{1 - 5}{4}

General solutions for

cos (2 x) = \frac{1 - 5}{4}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

2 x = arccos (\frac{1 - 5}{4}) + 2 πn, 2 x = - arccos (\frac{1 - 5}{4}) + 2 πn

2 x = arccos (\frac{1 - 5}{4}) + 2 πn, 2 x = - arccos (\frac{1 - 5}{4}) + 2 πn

Solve

2 x = arccos (\frac{1 - 5}{4}) + 2 πn : x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

2 x = arccos (\frac{1 - 5}{4}) + 2 πn

Divide both sides by

2

2 x = arccos (\frac{1 - 5}{4}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( \frac{1 - 5}{4} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

Solve

2 x = - arccos (\frac{1 - 5}{4}) + 2 πn : x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

2 x = - arccos (\frac{1 - 5}{4}) + 2 πn

Divide both sides by

2

2 x = - arccos (\frac{1 - 5}{4}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + \frac{2 πn}{2}

Simplify

x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn, x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

Combine all the solutions

x = \frac{π}{2} + πn, x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn, x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn, x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn, x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

Show solutions in decimal form

x = \frac{π}{2} + πn, x = \frac{0.62831 \dots}{2} + πn, x = π - \frac{0.62831 \dots}{2} + πn, x = \frac{1.88495 \dots}{2} + πn, x = - \frac{1.88495 \dots}{2} + πn

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