What is the general solution for 4cos^2(x)+cos(2x)-7cos(x)=-2 ?

The general solution for 4cos^2(x)+cos(2x)-7cos(x)=-2 is x=2pin,x=1.40334…+2pin,x=2pi-1.40334…+2pin

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4cos^2(x)+cos(2x)-7cos(x)=-2

Solution

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2

Solution

x = 2 πn, x = 1.40334 \dots + 2 πn, x = 2 π - 1.40334 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 80.40593 \dots^{\circ} + 36 0^{\circ} n, x = 279.59406 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2

Subtract

- 2

from both sides

4 cos^{2} (x) + cos (2 x) - 7 cos (x) + 2 = 0

Rewrite using trig identities

2 + cos (2 x) + 4 cos^{2} (x) - 7 cos (x)

Use the Double Angle identity:

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

= 2 + 2 cos^{2} (x) - 1 + 4 cos^{2} (x) - 7 cos (x)

Simplify

2 + 2 cos^{2} (x) - 1 + 4 cos^{2} (x) - 7 cos (x) : 6 cos^{2} (x) - 7 cos (x) + 1

2 + 2 cos^{2} (x) - 1 + 4 cos^{2} (x) - 7 cos (x)

Group like terms

= 2 cos^{2} (x) + 4 cos^{2} (x) - 7 cos (x) + 2 - 1

Add similar elements:

2 cos^{2} (x) + 4 cos^{2} (x) = 6 cos^{2} (x)

= 6 cos^{2} (x) - 7 cos (x) + 2 - 1

Add/Subtract the numbers:

2 - 1 = 1

= 6 cos^{2} (x) - 7 cos (x) + 1

= 6 cos^{2} (x) - 7 cos (x) + 1

1 + 6 cos^{2} (x) - 7 cos (x) = 0

Solve by substitution

1 + 6 cos^{2} (x) - 7 cos (x) = 0

Let:

cos (x) = u

1 + 6 u^{2} - 7 u = 0

1 + 6 u^{2} - 7 u = 0 : u = 1, u = \frac{1}{6}

1 + 6 u^{2} - 7 u = 0

Write in the standard form

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

6 u^{2} - 7 u + 1 = 0

Solve with the quadratic formula

6 u^{2} - 7 u + 1 = 0

Quadratic Equation Formula:

For

a = 6, b = - 7, c = 1

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 \cdot 1 = 5

(- 7)^{2} - 4 \cdot 6 \cdot 1

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 6}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 6}

u = \frac{- ( - 7 ) + 5}{2 \cdot 6} : 1

\frac{- ( - 7 ) + 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{7 + 5}{2 \cdot 6}

Add the numbers:

7 + 5 = 12

= \frac{12}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 7 ) - 5}{2 \cdot 6} : \frac{1}{6}

\frac{- ( - 7 ) - 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{7 - 5}{2 \cdot 6}

Subtract the numbers:

7 - 5 = 2

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{2}{12}

Cancel the common factor:

2

= \frac{1}{6}

The solutions to the quadratic equation are:

u = 1, u = \frac{1}{6}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = \frac{1}{6}

cos (x) = 1, cos (x) = \frac{1}{6}

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

cos (x) = \frac{1}{6} : x = arccos (\frac{1}{6}) + 2 πn, x = 2 π - arccos (\frac{1}{6}) + 2 πn

cos (x) = \frac{1}{6}

Apply trig inverse properties

cos (x) = \frac{1}{6}

General solutions for

cos (x) = \frac{1}{6}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{6}) + 2 πn, x = 2 π - arccos (\frac{1}{6}) + 2 πn

x = arccos (\frac{1}{6}) + 2 πn, x = 2 π - arccos (\frac{1}{6}) + 2 πn

Combine all the solutions

x = 2 πn, x = arccos (\frac{1}{6}) + 2 πn, x = 2 π - arccos (\frac{1}{6}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = 1.40334 \dots + 2 πn, x = 2 π - 1.40334 \dots + 2 πn

Popular Examples

4cos(x)=sqrt(2)+2cos(x)

4 cos (x) = 2 + 2 cos (x)

4cos(θ)=4sqrt(3)sin(θ)

4 cos (θ) = 43 sin (θ)

-4sin(θ)=4(sin(θ)-1)

- 4 sin (θ) = 4 (sin (θ) - 1)

4+2cos(θ)=0

4 + 2 cos (θ) = 0

9.81sin(θ)-5.88cos(θ)=2.77

9.81 sin (θ) - 5.88 cos (θ) = 2.77

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)+cos(2x)-7cos(x)=-2 ?
The general solution for 4cos^2(x)+cos(2x)-7cos(x)=-2 is x=2pin,x=1.40334…+2pin,x=2pi-1.40334…+2pin