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

The general solution for 4sin^2(x/2)+cos^2(x)=2 is x= pi/2+2pin,x=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

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

Solution

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

Solution

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

Degrees

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

Solution steps

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

Subtract

2

from both sides

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

Rewrite using trig identities

- 2 + 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}

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

Simplify

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

- 2 + 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)

= - 2 + 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

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

Simplify

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

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

Group like terms

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

- 2 + 2 = 0

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} - 2 u = 0

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

u^{2} - 2 u = 0

Solve with the quadratic formula

u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 0

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 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 = 2, u = 0

Substitute back

u = cos (x)

cos (x) = 2, cos (x) = 0

cos (x) = 2, cos (x) = 0

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

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

Graph

Popular Examples

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

12cot^2(x)=4

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

5sin^2(θ)+2sin(θ)=0

cos(θ)cos(3θ)-sin(θ)sin(3θ)=0

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x/2)+cos^2(x)=2 ?
The general solution for 4sin^2(x/2)+cos^2(x)=2 is x= pi/2+2pin,x=(3pi)/2+2pin