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

The general solution for 6cos^2(x)+cos(2x)=1 is x=(2pi)/3+pin,x= pi/3+pin

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

Solution

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

Solution

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

Degrees

x = 12 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

- 1 + cos (2 x) + 6 cos^{2} (x)

Use the Double Angle identity:

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

= - 1 + 1 - 2 sin^{2} (x) + 6 cos^{2} (x)

Simplify

= 6 cos^{2} (x) - 2 sin^{2} (x)

- 2 sin^{2} (x) + 6 cos^{2} (x) = 0

Factor

- 2 sin^{2} (x) + 6 cos^{2} (x) : 2 (3 cos (x) + sin (x)) (3 cos (x) - sin (x))

- 2 sin^{2} (x) + 6 cos^{2} (x)

Rewrite

6

3 \cdot 2

= - 2 sin^{2} (x) + 3 \cdot 2 cos^{2} (x)

Factor out common term

2

= 2 (- sin^{2} (x) + 3 cos^{2} (x))

Factor

3 cos^{2} (x) - sin^{2} (x) : (3 cos (x) + sin (x)) (3 cos (x) - sin (x))

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

Rewrite

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

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

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

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

= (3)^{2} cos^{2} (x) - sin^{2} (x)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(3)^{2} cos^{2} (x) = (3 cos (x))^{2}

= (3 cos (x))^{2} - sin^{2} (x)

= (3 cos (x))^{2} - sin^{2} (x)

Apply Difference of Two Squares Formula:

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

(3 cos (x))^{2} - sin^{2} (x) = (3 cos (x) + sin (x)) (3 cos (x) - sin (x))

= (3 cos (x) + sin (x)) (3 cos (x) - sin (x))

= 2 (3 cos (x) + sin (x)) (3 cos (x) - sin (x))

2 (3 cos (x) + sin (x)) (3 cos (x) - sin (x)) = 0

Solving each part separately

3 cos (x) + sin (x) = 0 or 3 cos (x) - sin (x) = 0

3 cos (x) + sin (x) = 0 : x = \frac{2 π}{3} + πn

3 cos (x) + sin (x) = 0

Rewrite using trig identities

3 cos (x) + sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{3 cos ( x ) + sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

3 + \frac{sin ( x )}{cos ( x )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

3 + tan (x) = 0

3 + tan (x) = 0

Move

3

to the right side

3 + tan (x) = 0

Subtract

3

from both sides

3 + tan (x) - 3 = 0 - 3

Simplify

tan (x) = - 3

tan (x) = - 3

General solutions for

tan (x) = - 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

3 cos (x) - sin (x) = 0 : x = \frac{π}{3} + πn

3 cos (x) - sin (x) = 0

Rewrite using trig identities

3 cos (x) - sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{3 cos ( x ) - sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

3 - tan (x) = 0

3 - tan (x) = 0

Move

3

to the right side

3 - tan (x) = 0

Subtract

3

from both sides

3 - tan (x) - 3 = 0 - 3

Simplify

- tan (x) = - 3

- tan (x) = - 3

Divide both sides by

- 1

- tan (x) = - 3

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 3}{- 1}

Simplify

tan (x) = 3

tan (x) = 3

General solutions for

tan (x) = 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

Combine all the solutions

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

Popular Examples

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sin (2 π t) = 1

(4cos^2(x)-3)(csc(x)+2)=0

(4 cos^{2} (x) - 3) (csc (x) + 2) = 0

sin(x)=cos(2x+10)

sin (x) = cos (2 x + 10)

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

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

6cos(x)=-6(1+cos(x))

6 cos (x) = - 6 (1 + cos (x))

Frequently Asked Questions (FAQ)

What is the general solution for 6cos^2(x)+cos(2x)=1 ?
The general solution for 6cos^2(x)+cos(2x)=1 is x=(2pi)/3+pin,x= pi/3+pin