What is the general solution for 4cos^2(x)-11cos(x)+6=0 ?

The general solution for 4cos^2(x)-11cos(x)+6=0 is x=0.72273…+2pin,x=2pi-0.72273…+2pin

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Popular Trigonometry >

4cos^2(x)-11cos(x)+6=0

Solution

4 cos^{2} (x) - 11 cos (x) + 6 = 0

Solution

x = 0.72273 \dots + 2 πn, x = 2 π - 0.72273 \dots + 2 πn

Degrees

x = 41.40962 \dots^{\circ} + 36 0^{\circ} n, x = 318.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) - 11 cos (x) + 6 = 0

Solve by substitution

4 cos^{2} (x) - 11 cos (x) + 6 = 0

Let:

cos (x) = u

4 u^{2} - 11 u + 6 = 0

4 u^{2} - 11 u + 6 = 0 : u = 2, u = \frac{3}{4}

4 u^{2} - 11 u + 6 = 0

Solve with the quadratic formula

4 u^{2} - 11 u + 6 = 0

Quadratic Equation Formula:

For

a = 4, b = - 11, c = 6

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 \cdot 6}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 \cdot 6}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 \cdot 6 = 5

(- 11)^{2} - 4 \cdot 4 \cdot 6

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 6 = 96

= 1 1^{2} - 96

1 1^{2} = 121

= 121 - 96

Subtract the numbers:

121 - 96 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 11 ) \pm 5}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 5}{2 \cdot 4}, u_{2} = \frac{- ( - 11 ) - 5}{2 \cdot 4}

u = \frac{- ( - 11 ) + 5}{2 \cdot 4} : 2

\frac{- ( - 11 ) + 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{11 + 5}{2 \cdot 4}

Add the numbers:

11 + 5 = 16

= \frac{16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= 2

u = \frac{- ( - 11 ) - 5}{2 \cdot 4} : \frac{3}{4}

\frac{- ( - 11 ) - 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{11 - 5}{2 \cdot 4}

Subtract the numbers:

11 - 5 = 6

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

The solutions to the quadratic equation are:

u = 2, u = \frac{3}{4}

Substitute back

u = cos (x)

cos (x) = 2, cos (x) = \frac{3}{4}

cos (x) = 2, cos (x) = \frac{3}{4}

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{3}{4} : x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

cos (x) = \frac{3}{4}

Apply trig inverse properties

cos (x) = \frac{3}{4}

General solutions for

cos (x) = \frac{3}{4}

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

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

Show solutions in decimal form

x = 0.72273 \dots + 2 πn, x = 2 π - 0.72273 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-11cos(x)+6=0 ?
The general solution for 4cos^2(x)-11cos(x)+6=0 is x=0.72273…+2pin,x=2pi-0.72273…+2pin