What is the general solution for 4cos^2(x)-11cos(x)+7=0 ?

The general solution for 4cos^2(x)-11cos(x)+7=0 is x=2pin

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Popular >

4cos^2(x)-11cos(x)+7=0

Solution

4 cos^{2} (x) - 11 cos (x) + 7 = 0

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) - 11 cos (x) + 7 = 0

Solve by substitution

4 cos^{2} (x) - 11 cos (x) + 7 = 0

Let:

cos (x) = u

4 u^{2} - 11 u + 7 = 0

4 u^{2} - 11 u + 7 = 0 : u = \frac{7}{4}, u = 1

4 u^{2} - 11 u + 7 = 0

Solve with the quadratic formula

4 u^{2} - 11 u + 7 = 0

Quadratic Equation Formula:

For

a = 4, b = - 11, c = 7

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 \cdot 7}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 \cdot 7}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 \cdot 7 = 3

(- 11)^{2} - 4 \cdot 4 \cdot 7

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 7 = 112

= 1 1^{2} - 112

1 1^{2} = 121

= 121 - 112

Subtract the numbers:

121 - 112 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 4}, u_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 4}

u = \frac{- ( - 11 ) + 3}{2 \cdot 4} : \frac{7}{4}

\frac{- ( - 11 ) + 3}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{11 + 3}{2 \cdot 4}

Add the numbers:

11 + 3 = 14

= \frac{14}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{14}{8}

Cancel the common factor:

2

= \frac{7}{4}

u = \frac{- ( - 11 ) - 3}{2 \cdot 4} : 1

\frac{- ( - 11 ) - 3}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{11 - 3}{2 \cdot 4}

Subtract the numbers:

11 - 3 = 8

= \frac{8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{7}{4}, u = 1

Substitute back

u = cos (x)

cos (x) = \frac{7}{4}, cos (x) = 1

cos (x) = \frac{7}{4}, cos (x) = 1

cos (x) = \frac{7}{4} :

No Solution

cos (x) = \frac{7}{4}

- 1 \leq cos (x) \leq 1

No Solution

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

Combine all the solutions

x = 2 πn

Popular Examples

sqrt(3)tan(x)+2sec(x)=1

3 tan (x) + 2 sec (x) = 1

cos(a)=0.14

cos (a) = 0.14

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

\frac{1 + cos ^{2} ( x )}{sin ^{2} ( x )} = \frac{1}{sin ^{2} ( x )}

7sin(a)= 1/11

7 sin (a) = \frac{1}{11}

tan(x)=2,x+y=135

tan (x) = 2, x + y = 135

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-11cos(x)+7=0 ?
The general solution for 4cos^2(x)-11cos(x)+7=0 is x=2pin