What is the general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 ?

Q: What is the general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 ?

The general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 is x=1.23095…,x=360-1.23095…,x=120,x=240

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6cos^2(x)+cos(x)=1,0<= x<= 360

Solution

6 cos^{2} (x) + cos (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 1.23095 \dots, x = 36 0^{\circ} - 1.23095 \dots, x = 12 0^{\circ}, x = 24 0^{\circ}

Radians

x = 1.23095 \dots, x = 2 π - 1.23095 \dots, x = \frac{2 π}{3}, x = \frac{4 π}{3}

Solution steps

6 cos^{2} (x) + cos (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

Solve by substitution

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

Let:

cos (x) = u

6 u^{2} + u = 1

6 u^{2} + u = 1 : u = \frac{1}{3}, u = - \frac{1}{2}

6 u^{2} + u = 1

Move

1

to the left side

6 u^{2} + u = 1

Subtract

1

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 6, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

1^{2} - 4 \cdot 6 (- 1) = 5

1^{2} - 4 \cdot 6 (- 1)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 6 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 6}, u_{2} = \frac{- 1 - 5}{2 \cdot 6}

u = \frac{- 1 + 5}{2 \cdot 6} : \frac{1}{3}

\frac{- 1 + 5}{2 \cdot 6}

Add/Subtract the numbers:

- 1 + 5 = 4

= \frac{4}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{4}{12}

Cancel the common factor:

4

= \frac{1}{3}

u = \frac{- 1 - 5}{2 \cdot 6} : - \frac{1}{2}

\frac{- 1 - 5}{2 \cdot 6}

Subtract the numbers:

- 1 - 5 = - 6

= \frac{- 6}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{1}{3}, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1}{3}, cos (x) = - \frac{1}{2}

cos (x) = \frac{1}{3}, cos (x) = - \frac{1}{2}

cos (x) = \frac{1}{3}, 0 \leq x \leq 36 0^{\circ} : x = arccos (\frac{1}{3}), x = 36 0^{\circ} - arccos (\frac{1}{3})

cos (x) = \frac{1}{3}, 0 \leq x \leq 36 0^{\circ}

Apply trig inverse properties

cos (x) = \frac{1}{3}

General solutions for

cos (x) = \frac{1}{3}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

x = arccos (\frac{1}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1}{3}) + 36 0^{\circ} n

x = arccos (\frac{1}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1}{3}) + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = arccos (\frac{1}{3}), x = 36 0^{\circ} - arccos (\frac{1}{3})

cos (x) = - \frac{1}{2}, 0 \leq x \leq 36 0^{\circ} : x = 12 0^{\circ}, x = 24 0^{\circ}

cos (x) = - \frac{1}{2}, 0 \leq x \leq 36 0^{\circ}

General solutions for

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

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 12 0^{\circ}, x = 24 0^{\circ}

Combine all the solutions

x = arccos (\frac{1}{3}), x = 36 0^{\circ} - arccos (\frac{1}{3}), x = 12 0^{\circ}, x = 24 0^{\circ}

Show solutions in decimal form

x = 1.23095 \dots, x = 36 0^{\circ} - 1.23095 \dots, x = 12 0^{\circ}, x = 24 0^{\circ}

Popular Examples

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tan (y) = \frac{π}{4}

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2 sin (x - 1) + 1 = 0

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csc(x)+2=0,0<= x<= 2pi

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tan(x)=3.7

tan (x) = 3.7

Frequently Asked Questions (FAQ)

What is the general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 ?
The general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 is x=1.23095…,x=360-1.23095…,x=120,x=240