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

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

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

Solution

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

Solution

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

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Rewrite using trig identities

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)

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

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

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

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

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

Apply rule

- (- a) = a

= (- 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

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a, - (- 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 the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Remove parentheses:

(- a) = - a, - (- 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 the fraction rule:

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

= - \frac{0}{2}

Apply rule

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

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

Popular Examples

-cos(a)-11=2cos(a)-9

- cos (a) - 11 = 2 cos (a) - 9

4-2sin(x)=0

4 - 2 sin (x) = 0

sin^2(θ)-2sin(θ)+3=0

sin^{2} (θ) - 2 sin (θ) + 3 = 0

3tan^2(θ)+sqrt(3)tan(θ)=0,0<= θ<= 360

3 tan^{2} (θ) + 3 tan (θ) = 0, 0^{\circ} \leq θ \leq 36 0^{\circ}

cos(a)= 24/25

cos (a) = \frac{24}{25}

Frequently Asked Questions (FAQ)

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