What is the general solution for 6cos^2(x/2)+cos(x)+1=0 ?

The general solution for 6cos^2(x/2)+cos(x)+1=0 is x=pi+2pin

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6cos^2(x/2)+cos(x)+1=0

Solution

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

Solution

x = π + 2 πn

Degrees

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

Solution steps

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

Rewrite using trig identities

1 + cos (x) + 6 cos^{2} (\frac{x}{2})

Use the following identity:

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

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Switch sides

2 cos^{2} (θ) - 1 = cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 + cos (2 θ)

Divide both sides by

2

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

= 1 + cos (x) + 6 \cdot \frac{1 + cos ( 2 \cdot \frac{x}{2} )}{2}

Simplify

1 + cos (x) + 6 \cdot \frac{1 + cos ( 2 \cdot \frac{x}{2} )}{2} : 4 cos (x) + 4

1 + cos (x) + 6 \cdot \frac{1 + cos ( 2 \cdot \frac{x}{2} )}{2}

6 \cdot \frac{1 + cos ( 2 \cdot \frac{x}{2} )}{2} = 3 (cos (x) + 1)

6 \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}

= 6 \cdot \frac{cos ( x ) + 1}{2}

Multiply fractions:

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

= \frac{( 1 + cos ( x ) ) \cdot 6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 (cos (x) + 1)

= 1 + cos (x) + 3 (cos (x) + 1)

Expand

3 (cos (x) + 1) : 3 cos (x) + 3

3 (cos (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = cos (x), c = 1

= 3 cos (x) + 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 3 cos (x) + 3

= 1 + cos (x) + 3 cos (x) + 3

Simplify

1 + cos (x) + 3 cos (x) + 3 : 4 cos (x) + 4

1 + cos (x) + 3 cos (x) + 3

Add similar elements:

cos (x) + 3 cos (x) = 4 cos (x)

= 1 + 4 cos (x) + 3

Group like terms

= 4 cos (x) + 1 + 3

Add the numbers:

1 + 3 = 4

= 4 cos (x) + 4

= 4 cos (x) + 4

= 4 cos (x) + 4

4 + 4 cos (x) = 0

Move

4

to the right side

4 + 4 cos (x) = 0

Subtract

4

from both sides

4 + 4 cos (x) - 4 = 0 - 4

Simplify

4 cos (x) = - 4

4 cos (x) = - 4

Divide both sides by

4

4 cos (x) = - 4

Divide both sides by

4

\frac{4 cos ( x )}{4} = \frac{- 4}{4}

Simplify

cos (x) = - 1

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

Popular Examples

-2sin(θ)=1

- 2 sin (θ) = 1

cos(θ)=-7/3

cos (θ) = - \frac{7}{3}

0=-6sin(x)+6cos(2x)

0 = - 6 sin (x) + 6 cos (2 x)

tan(2x+1)= 1/2

tan (2 x + 1) = \frac{1}{2}

cos(B)= 8/12

cos (B) = \frac{8}{12}

Frequently Asked Questions (FAQ)

What is the general solution for 6cos^2(x/2)+cos(x)+1=0 ?
The general solution for 6cos^2(x/2)+cos(x)+1=0 is x=pi+2pin