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

The general solution for 2-2cos^2(x/2)=sin^2(x) is x=2pin,x= pi/2+2pin,x=(3pi)/2+2pin

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

Solution

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

Solution

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

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

2 - sin^{2} (x) - 2 cos^{2} (\frac{x}{2})

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Subtract the numbers:

2 - 1 = 1

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

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

Use the Double Angle identity:

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

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

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

Multiply

2 \cdot \frac{x}{2} : x

2 \cdot \frac{x}{2}

Multiply fractions:

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

= \frac{x \cdot 2}{2}

Cancel the common factor:

2

= x

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

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

Solve by substitution

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

Let:

cos (x) = u

- u + u^{2} = 0

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

- u + u^{2} = 0

Write in the standard form

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

u^{2} - u = 0

Solve with the quadratic formula

u^{2} - u = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 1 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 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 ) - 1}{2 \cdot 1} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 1}

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = 0

cos (x) = 1, cos (x) = 0

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) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2-2cos^2(x/2)=sin^2(x) ?
The general solution for 2-2cos^2(x/2)=sin^2(x) is x=2pin,x= pi/2+2pin,x=(3pi)/2+2pin