What is the general solution for cos^4(x)-2sin^2(x)-1=0 ?

The general solution for cos^4(x)-2sin^2(x)-1=0 is x=2pin,x=pi+2pin

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

- 2 (1 - cos^{2} (x)) : - 2 + 2 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = cos^{2} (x)

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 1 - 2 = - 3

= cos^{4} (x) + 2 cos^{2} (x) - 3

= cos^{4} (x) + 2 cos^{2} (x) - 3

= cos^{4} (x) + 2 cos^{2} (x) - 3

- 3 + cos^{4} (x) + 2 cos^{2} (x) = 0

Solve by substitution

- 3 + cos^{4} (x) + 2 cos^{2} (x) = 0

Let:

cos (x) = u

- 3 + u^{4} + 2 u^{2} = 0

- 3 + u^{4} + 2 u^{2} = 0 : u = 1, u = - 1, u = 3 i, u = - 3 i

- 3 + u^{4} + 2 u^{2} = 0

Write in the standard form

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

u^{4} + 2 u^{2} - 3 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} + 2 v - 3 = 0

Solve

v^{2} + 2 v - 3 = 0 : v = 1, v = - 3

v^{2} + 2 v - 3 = 0

Solve with the quadratic formula

v^{2} + 2 v - 3 = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = - 3

v_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

v_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 3) = 4

2^{2} - 4 \cdot 1 \cdot (- 3)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

v_{1} = \frac{- 2 + 4}{2 \cdot 1}, v_{2} = \frac{- 2 - 4}{2 \cdot 1}

v = \frac{- 2 + 4}{2 \cdot 1} : 1

\frac{- 2 + 4}{2 \cdot 1}

Add/Subtract the numbers:

- 2 + 4 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

v = \frac{- 2 - 4}{2 \cdot 1} : - 3

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

Subtract the numbers:

- 2 - 4 = - 6

= \frac{- 6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6}{2}

Apply the fraction rule:

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

= - \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3

The solutions to the quadratic equation are:

v = 1, v = - 3

v = 1, v = - 3

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Solve

u^{2} = - 3 : u = 3 i, u = - 3 i

u^{2} = - 3

For

x^{2} = f (a)

the solutions are

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

u = - 3, u = - - 3

Simplify

- 3 : 3 i

- 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

Simplify

- - 3 : - 3 i

- - 3

Simplify

- 3 : 3 i

- 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

= - 3 i

u = 3 i, u = - 3 i

The solutions are

u = 1, u = - 1, u = 3 i, u = - 3 i

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - 1, cos (x) = 3 i, cos (x) = - 3 i

cos (x) = 1, cos (x) = - 1, cos (x) = 3 i, cos (x) = - 3 i

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) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

cos (x) = 3 i :

No Solution

cos (x) = 3 i

No Solution

cos (x) = - 3 i :

No Solution

cos (x) = - 3 i

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

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cos^4(x)-2cos^2(x)+1=0

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Frequently Asked Questions (FAQ)

What is the general solution for cos^4(x)-2sin^2(x)-1=0 ?
The general solution for cos^4(x)-2sin^2(x)-1=0 is x=2pin,x=pi+2pin