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

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

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

Solution

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

Solution

x = πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 1 - 1 = - 2

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

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

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

Solve by substitution

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

Let:

cos (2 x) = u

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

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

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

Write in the standard form

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

u^{2} + u - 2 = 0

Solve with the quadratic formula

u^{2} + u - 2 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 \cdot 1}, u_{2} = \frac{- 1 - 3}{2 \cdot 1}

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

\frac{- 1 + 3}{2 \cdot 1}

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

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

\frac{- 1 - 3}{2 \cdot 1}

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = 1, u = - 2

Substitute back

u = cos (2 x)

cos (2 x) = 1, cos (2 x) = - 2

cos (2 x) = 1, cos (2 x) = - 2

cos (2 x) = 1 : x = 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 = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

cos (2 x) = - 2 :

No Solution

cos (2 x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = πn

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Frequently Asked Questions (FAQ)

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