What is the general solution for 1-2cos^2(8x)=sin(4x) ?

The general solution for 1-2cos^2(8x)=sin(4x) is

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1-2cos^2(8x)=sin(4x)

Solution

1 - 2 cos^{2} (8 x) = sin (4 x)

Solution

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{0.94247 \dots + 2 πn}{4}, x = \frac{π - 0.94247 \dots + 2 πn}{4}, x = \frac{- 0.31415 \dots + 2 πn}{4}, x = \frac{π + 0.31415 \dots + 2 πn}{4}

Degrees

x = 67. 5^{\circ} + 9 0^{\circ} n, x = 7. 5^{\circ} + 9 0^{\circ} n, x = 37. 5^{\circ} + 9 0^{\circ} n, x = 13. 5^{\circ} + 9 0^{\circ} n, x = 31. 5^{\circ} + 9 0^{\circ} n, x = - 4. 5^{\circ} + 9 0^{\circ} n, x = 49. 5^{\circ} + 9 0^{\circ} n

Solution steps

1 - 2 cos^{2} (8 x) = sin (4 x)

Subtract

sin (4 x)

from both sides

1 - 2 cos^{2} (8 x) - sin (4 x) = 0

Let:

u = 4 x

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

1 - sin (u) - 2 (1 - 2 sin^{2} (u))^{2} : 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

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

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

Apply Perfect Square Formula:

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

a = 1, b = 2 sin^{2} (u)

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 2 sin^{2} (u) = 4 sin^{2} (u)

2 \cdot 1 \cdot 2 sin^{2} (u)

Multiply the numbers:

2 \cdot 1 \cdot 2 = 4

= 4 sin^{2} (u)

(2 sin^{2} (u))^{2} = 4 sin^{4} (u)

(2 sin^{2} (u))^{2}

Apply exponent rule:

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

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

(sin^{2} (u))^{2} : sin^{4} (u)

Apply exponent rule:

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

= sin^{2 \cdot 2} (u)

Multiply the numbers:

2 \cdot 2 = 4

= sin^{4} (u)

= 2^{2} sin^{4} (u)

2^{2} = 4

= 4 sin^{4} (u)

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

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

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

Expand

- 2 (1 - 4 sin^{2} (u) + 4 sin^{4} (u)) : - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

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

Simplify

- 2 \cdot 1 + 2 \cdot 4 sin^{2} (u) - 2 \cdot 4 sin^{4} (u) : - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

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

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

= - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

= 1 - sin (u) - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

Simplify

1 - sin (u) - 2 + 8 sin^{2} (u) - 8 sin^{4} (u) : 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

1 - sin (u) - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

Group like terms

= - sin (u) + 8 sin^{2} (u) - 8 sin^{4} (u) + 1 - 2

Add/Subtract the numbers:

1 - 2 = - 1

= 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

= 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

= 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

- 1 - sin (u) + 8 sin^{2} (u) - 8 sin^{4} (u) = 0

Solve by substitution

- 1 - sin (u) + 8 sin^{2} (u) - 8 sin^{4} (u) = 0

Let:

sin (u) = u

- 1 - u + 8 u^{2} - 8 u^{4} = 0

- 1 - u + 8 u^{2} - 8 u^{4} = 0 : u = - 1, u = \frac{1}{2}, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

- 1 - u + 8 u^{2} - 8 u^{4} = 0

Write in the standard form

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

- 8 u^{4} + 8 u^{2} - u - 1 = 0

Factor

- 8 u^{4} + 8 u^{2} - u - 1 : - (u + 1) (2 u - 1) (4 u^{2} - 2 u - 1)

- 8 u^{4} + 8 u^{2} - u - 1

Factor out common term

- 1

= - (8 u^{4} - 8 u^{2} + u + 1)

Factor

8 u^{4} - 8 u^{2} + u + 1 : (u + 1) (2 u - 1) (4 u^{2} - 2 u - 1)

8 u^{4} - 8 u^{2} + u + 1

Use the rational root theorem

a_{0} = 1, a_{n} = 8

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4, 8

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4 , 8}

- \frac{1}{1}

is a root of the expression, so factor out

u + 1

= (u + 1) \frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1}

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1} = 8 u^{3} - 8 u^{2} + 1

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1}

Divide

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1} : \frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1} = 8 u^{3} + \frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1}

Divide the leading coefficients of the numerator

8 u^{4} - 8 u^{2} + u + 1

and the divisor

u + 1 : \frac{8 u ^{4}}{u} = 8 u^{3}

Quotient = 8 u^{3}

Multiply

u + 1

8 u^{3} : 8 u^{4} + 8 u^{3}

Subtract

8 u^{4} + 8 u^{3}

from

8 u^{4} - 8 u^{2} + u + 1

to get new remainder

Remainder = - 8 u^{3} - 8 u^{2} + u + 1

Therefore

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1} = 8 u^{3} + \frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1}

= 8 u^{3} + \frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1}

Divide

\frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1} : \frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1} = - 8 u^{2} + \frac{u + 1}{u + 1}

Divide the leading coefficients of the numerator

- 8 u^{3} - 8 u^{2} + u + 1

and the divisor

u + 1 : \frac{- 8 u ^{3}}{u} = - 8 u^{2}

Quotient = - 8 u^{2}

Multiply

u + 1

- 8 u^{2} : - 8 u^{3} - 8 u^{2}

Subtract

- 8 u^{3} - 8 u^{2}

from

- 8 u^{3} - 8 u^{2} + u + 1

to get new remainder

Remainder = u + 1

Therefore

\frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1} = - 8 u^{2} + \frac{u + 1}{u + 1}

= 8 u^{3} - 8 u^{2} + \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

= 8 u^{3} - 8 u^{2} + 1

= 8 u^{3} - 8 u^{2} + 1

Factor

8 u^{3} - 8 u^{2} + 1 : (2 u - 1) (4 u^{2} - 2 u - 1)

8 u^{3} - 8 u^{2} + 1

Use the rational root theorem

a_{0} = 1, a_{n} = 8

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4, 8

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4 , 8}

\frac{1}{2}

is a root of the expression, so factor out

2 u - 1

= (2 u - 1) \frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1}

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1} = 4 u^{2} - 2 u - 1

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1}

Divide

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1} : \frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1} = 4 u^{2} + \frac{- 4 u ^{2} + 1}{2 u - 1}

Divide the leading coefficients of the numerator

8 u^{3} - 8 u^{2} + 1

and the divisor

2 u - 1 : \frac{8 u ^{3}}{2 u} = 4 u^{2}

Quotient = 4 u^{2}

Multiply

2 u - 1

4 u^{2} : 8 u^{3} - 4 u^{2}

Subtract

8 u^{3} - 4 u^{2}

from

8 u^{3} - 8 u^{2} + 1

to get new remainder

Remainder = - 4 u^{2} + 1

Therefore

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1} = 4 u^{2} + \frac{- 4 u ^{2} + 1}{2 u - 1}

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

Divide

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

Divide the leading coefficients of the numerator

- 4 u^{2} + 1

and the divisor

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

Quotient = - 2 u

Multiply

2 u - 1

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

Subtract

- 4 u^{2} + 2 u

from

- 4 u^{2} + 1

to get new remainder

Remainder = - 2 u + 1

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

- 2 u + 1

and the divisor

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

Quotient = - 1

Multiply

2 u - 1

- 1 : - 2 u + 1

Subtract

- 2 u + 1

from

- 2 u + 1

to get new remainder

Remainder = 0

Therefore

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

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

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

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

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

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

- (u + 1) (2 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 2 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

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

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

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

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

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

Substitute back

u = sin (u)

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

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

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

sin (u) = - 1

General solutions for

sin (u) = - 1

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}

u = \frac{3 π}{2} + 2 πn

u = \frac{3 π}{2} + 2 πn

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

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

General solutions for

sin (u) = \frac{1}{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}

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

sin (u) = \frac{1 + 5}{4} : u = arcsin (\frac{1 + 5}{4}) + 2 πn, u = π - arcsin (\frac{1 + 5}{4}) + 2 πn

sin (u) = \frac{1 + 5}{4}

Apply trig inverse properties

sin (u) = \frac{1 + 5}{4}

General solutions for

sin (u) = \frac{1 + 5}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

sin (u) = \frac{1 - 5}{4} : u = arcsin (\frac{1 - 5}{4}) + 2 πn, u = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

sin (u) = \frac{1 - 5}{4}

Apply trig inverse properties

sin (u) = \frac{1 - 5}{4}

General solutions for

sin (u) = \frac{1 - 5}{4}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

Combine all the solutions

u = \frac{3 π}{2} + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = arcsin (\frac{1 + 5}{4}) + 2 πn, u = π - arcsin (\frac{1 + 5}{4}) + 2 πn, u = arcsin (\frac{1 - 5}{4}) + 2 πn, u = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

Substitute back

u = 4 x

4 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π + 4 πn}{8}

4 x = \frac{3 π}{2} + 2 πn

Divide both sides by

4

4 x = \frac{3 π}{2} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} : \frac{3 π + 4 πn}{8}

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4}

Apply rule

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

= \frac{\frac{3 π}{2} + 2 πn}{4}

Join

\frac{3 π}{2} + 2 πn : \frac{3 π + 4 πn}{2}

\frac{3 π}{2} + 2 πn

Convert element to fraction:

2 πn = \frac{2 πn 2}{2}

= \frac{3 π}{2} + \frac{2 πn \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{3 π + 2 πn \cdot 2}{2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π + 4 πn}{2}

= \frac{\frac{3 π + 4 πn}{2}}{4}

Apply the fraction rule:

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

= \frac{3 π + 4 πn}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

4 x = \frac{π}{6} + 2 πn : x = \frac{π + 12 πn}{24}

4 x = \frac{π}{6} + 2 πn

Divide both sides by

4

4 x = \frac{π}{6} + 2 πn

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : \frac{π + 12 πn}{24}

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

Apply rule

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

= \frac{\frac{π}{6} + 2 πn}{4}

Join

\frac{π}{6} + 2 πn : \frac{π + 12 πn}{6}

\frac{π}{6} + 2 πn

Convert element to fraction:

2 πn = \frac{2 πn 6}{6}

= \frac{π}{6} + \frac{2 πn \cdot 6}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 2 πn \cdot 6}{6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{π + 12 πn}{6}

= \frac{\frac{π + 12 πn}{6}}{4}

Apply the fraction rule:

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

= \frac{π + 12 πn}{6 \cdot 4}

Multiply the numbers:

6 \cdot 4 = 24

= \frac{π + 12 πn}{24}

x = \frac{π + 12 πn}{24}

x = \frac{π + 12 πn}{24}

x = \frac{π + 12 πn}{24}

4 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π + 12 πn}{24}

4 x = \frac{5 π}{6} + 2 πn

Divide both sides by

4

4 x = \frac{5 π}{6} + 2 πn

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4} : \frac{5 π + 12 πn}{24}

\frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

Apply rule

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

= \frac{\frac{5 π}{6} + 2 πn}{4}

Join

\frac{5 π}{6} + 2 πn : \frac{5 π + 12 πn}{6}

\frac{5 π}{6} + 2 πn

Convert element to fraction:

2 πn = \frac{2 πn 6}{6}

= \frac{5 π}{6} + \frac{2 πn \cdot 6}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{5 π + 2 πn \cdot 6}{6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{5 π + 12 πn}{6}

= \frac{\frac{5 π + 12 πn}{6}}{4}

Apply the fraction rule:

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

= \frac{5 π + 12 πn}{6 \cdot 4}

Multiply the numbers:

6 \cdot 4 = 24

= \frac{5 π + 12 πn}{24}

x = \frac{5 π + 12 πn}{24}

x = \frac{5 π + 12 πn}{24}

x = \frac{5 π + 12 πn}{24}

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

4 x = arcsin (\frac{1 + 5}{4}) + 2 πn

Divide both sides by

4

4 x = arcsin (\frac{1 + 5}{4}) + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{arcsin ( \frac{1 + 5}{4} )}{4} + \frac{2 πn}{4} : \frac{arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}

\frac{arcsin ( \frac{1 + 5}{4} )}{4} + \frac{2 πn}{4}

Apply rule

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

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

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

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

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

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

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

Divide both sides by

4

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

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{π}{4} - \frac{arcsin ( \frac{1 + 5}{4} )}{4} + \frac{2 πn}{4} : \frac{π - arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}

\frac{π}{4} - \frac{arcsin ( \frac{1 + 5}{4} )}{4} + \frac{2 πn}{4}

Apply rule

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

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

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

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

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

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

4 x = arcsin (\frac{1 - 5}{4}) + 2 πn

Divide both sides by

4

4 x = arcsin (\frac{1 - 5}{4}) + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{arcsin ( \frac{1 - 5}{4} )}{4} + \frac{2 πn}{4} : \frac{arcsin ( \frac{1 - 5}{4} ) + 2 πn}{4}

\frac{arcsin ( \frac{1 - 5}{4} )}{4} + \frac{2 πn}{4}

Apply rule

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

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

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

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

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

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

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

- \frac{1 - 5}{4} = - \frac{- ( 5 - 1 )}{4} = \frac{5 - 1}{4}

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

Divide both sides by

4

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

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{π}{4} + \frac{arcsin ( \frac{5 - 1}{4} )}{4} + \frac{2 πn}{4} : \frac{π + arcsin ( \frac{5 - 1}{4} ) + 2 πn}{4}

\frac{π}{4} + \frac{arcsin ( \frac{5 - 1}{4} )}{4} + \frac{2 πn}{4}

Apply rule

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

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

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

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

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

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}, x = \frac{π - arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}, x = \frac{arcsin ( \frac{1 - 5}{4} ) + 2 πn}{4}, x = \frac{π + arcsin ( \frac{5 - 1}{4} ) + 2 πn}{4}

Show solutions in decimal form

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{0.94247 \dots + 2 πn}{4}, x = \frac{π - 0.94247 \dots + 2 πn}{4}, x = \frac{- 0.31415 \dots + 2 πn}{4}, x = \frac{π + 0.31415 \dots + 2 πn}{4}

Popular Examples

sin^2(x)=cos(2x)

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

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

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

tan(x-10)=0

tan (x - 1 0^{\circ}) = 0

cos^2(x)=3sin(x)cos(x)

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

cos(θ)-3cos(2θ)=0

cos (θ) - 3 cos (2 θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 1-2cos^2(8x)=sin(4x) ?
The general solution for 1-2cos^2(8x)=sin(4x) is