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

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

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

Solution

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

Solution

x = 4 πn, x = 2 π + 4 πn

Degrees

x = 0^{\circ} + 72 0^{\circ} n, x = 36 0^{\circ} + 72 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Simplify

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Add similar elements:

cos (x) - cos (x) = 0

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

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

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

Solve by substitution

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

Let:

cos (\frac{x}{2}) = u

- 1 + u^{2} = 0

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

- 1 + u^{2} = 0

Move

1

to the right side

- 1 + u^{2} = 0

Add

1

to both sides

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

Simplify

u^{2} = 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

Substitute back

u = cos (\frac{x}{2})

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

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

cos (\frac{x}{2}) = 1 : x = 4 πn

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

General solutions for

cos (\frac{x}{2}) = 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}

\frac{x}{2} = 0 + 2 πn

\frac{x}{2} = 0 + 2 πn

Solve

\frac{x}{2} = 0 + 2 πn : x = 4 πn

\frac{x}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{x}{2} = 2 πn

Multiply both sides by

2

\frac{x}{2} = 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot 2 πn

Simplify

x = 4 πn

x = 4 πn

x = 4 πn

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

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

General solutions for

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

\frac{x}{2} = π + 2 πn

\frac{x}{2} = π + 2 πn

Solve

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

\frac{x}{2} = π + 2 πn

Multiply both sides by

2

\frac{x}{2} = π + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 π + 2 \cdot 2 πn

Simplify

x = 2 π + 4 πn

x = 2 π + 4 πn

x = 2 π + 4 πn

Combine all the solutions

x = 4 πn, x = 2 π + 4 πn

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sin^{2} (a) = \frac{( 2 tan ( a ) )}{( 1 + tan ^{2} ( a ) )}

(tan(a))/2 = 2/11

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Frequently Asked Questions (FAQ)

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