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

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

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cos^2(x)+4sin^2(x/2)=1

Solution

cos^{2} (x) + 4 sin^{2} (\frac{x}{2}) = 1

Solution

x = 2 πn

Degrees

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

Solution steps

cos^{2} (x) + 4 sin^{2} (\frac{x}{2}) = 1

Subtract

1

from both sides

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

Rewrite using trig identities

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

Use the following identity:

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

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Switch sides

2 sin^{2} (θ) - 1 = - cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 - cos (2 θ)

Divide both sides by

2

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

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

Simplify

- 1 + cos^{2} (x) + 4 \cdot \frac{1 - cos ( 2 \cdot \frac{x}{2} )}{2} : cos^{2} (x) - 2 cos (x) + 1

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

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

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

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

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

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

= 4 \cdot \frac{- cos ( x ) + 1}{2}

Multiply fractions:

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

= \frac{( 1 - cos ( x ) ) \cdot 4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 (- cos (x) + 1)

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

Expand

2 (- cos (x) + 1) : - 2 cos (x) + 2

2 (- cos (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

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

= 2 (- cos (x)) + 2 \cdot 1

Apply minus-plus rules

+ (- a) = - a

= - 2 cos (x) + 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= - 2 cos (x) + 2

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

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

Combine all the solutions

x = 2 πn

Popular Examples

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

cos^2(x)+sin^2(x)-2sin(x)=0,0<= x<= 2pi

cos^{2} (x) + sin^{2} (x) - 2 sin (x) = 0, 0 \leq x \leq 2 π

cot(A)= 24/7

cot (A) = \frac{24}{7}

2sqrt(3)sec(θ)+4=0

23 sec (θ) + 4 = 0

sqrt(3)sin(x)=2

3 sin (x) = 2

Frequently Asked Questions (FAQ)

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