What is the general solution for 2cos^2(x)=11cos(x)+1 ?

The general solution for 2cos^2(x)=11cos(x)+1 is x=1.66037…+2pin,x=-1.66037…+2pin

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2cos^2(x)=11cos(x)+1

Solution

2 cos^{2} (x) = 11 cos (x) + 1

Solution

x = 1.66037 \dots + 2 πn, x = - 1.66037 \dots + 2 πn

Degrees

x = 95.13220 \dots^{\circ} + 36 0^{\circ} n, x = - 95.13220 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (x) = 11 cos (x) + 1

Solve by substitution

2 cos^{2} (x) = 11 cos (x) + 1

Let:

cos (x) = u

2 u^{2} = 11 u + 1

2 u^{2} = 11 u + 1 : u = \frac{11 + 129}{4}, u = \frac{11 - 129}{4}

2 u^{2} = 11 u + 1

Move

1

to the left side

2 u^{2} = 11 u + 1

Subtract

1

from both sides

2 u^{2} - 1 = 11 u + 1 - 1

Simplify

2 u^{2} - 1 = 11 u

2 u^{2} - 1 = 11 u

Move

11 u

to the left side

2 u^{2} - 1 = 11 u

Subtract

11 u

from both sides

2 u^{2} - 1 - 11 u = 11 u - 11 u

Simplify

2 u^{2} - 1 - 11 u = 0

2 u^{2} - 1 - 11 u = 0

Write in the standard form

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

2 u^{2} - 11 u - 1 = 0

Solve with the quadratic formula

2 u^{2} - 11 u - 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 11, c = - 1

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 (- 1) = 129

(- 11)^{2} - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= (- 11)^{2} + 4 \cdot 2 \cdot 1

Apply exponent rule:

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

n

is even

(- 11)^{2} = 1 1^{2}

= 1 1^{2} + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 1^{2} + 8

1 1^{2} = 121

= 121 + 8

Add the numbers:

121 + 8 = 129

= 129

u_{1, 2} = \frac{- ( - 11 ) \pm 129}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 129}{2 \cdot 2}, u_{2} = \frac{- ( - 11 ) - 129}{2 \cdot 2}

u = \frac{- ( - 11 ) + 129}{2 \cdot 2} : \frac{11 + 129}{4}

\frac{- ( - 11 ) + 129}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{11 + 129}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{11 + 129}{4}

u = \frac{- ( - 11 ) - 129}{2 \cdot 2} : \frac{11 - 129}{4}

\frac{- ( - 11 ) - 129}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{11 - 129}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{11 - 129}{4}

The solutions to the quadratic equation are:

u = \frac{11 + 129}{4}, u = \frac{11 - 129}{4}

Substitute back

u = cos (x)

cos (x) = \frac{11 + 129}{4}, cos (x) = \frac{11 - 129}{4}

cos (x) = \frac{11 + 129}{4}, cos (x) = \frac{11 - 129}{4}

cos (x) = \frac{11 + 129}{4} :

No Solution

cos (x) = \frac{11 + 129}{4}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{11 - 129}{4} : x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

cos (x) = \frac{11 - 129}{4}

Apply trig inverse properties

cos (x) = \frac{11 - 129}{4}

General solutions for

cos (x) = \frac{11 - 129}{4}

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

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

Show solutions in decimal form

x = 1.66037 \dots + 2 πn, x = - 1.66037 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)=11cos(x)+1 ?
The general solution for 2cos^2(x)=11cos(x)+1 is x=1.66037…+2pin,x=-1.66037…+2pin