What is the general solution for cos(2x)-7cos(x)+6=0 ?

The general solution for cos(2x)-7cos(x)+6=0 is x=2pin

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cos(2x)-7cos(x)+6=0

Solution

cos (2 x) - 7 cos (x) + 6 = 0

Solution

x = 2 πn

Degrees

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

Solution steps

cos (2 x) - 7 cos (x) + 6 = 0

Rewrite using trig identities

6 + cos (2 x) - 7 cos (x)

Use the Double Angle identity:

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

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

Simplify

6 + 2 cos^{2} (x) - 1 - 7 cos (x) : 2 cos^{2} (x) - 7 cos (x) + 5

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

Group like terms

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

Add/Subtract the numbers:

6 - 1 = 5

= 2 cos^{2} (x) - 7 cos (x) + 5

= 2 cos^{2} (x) - 7 cos (x) + 5

5 + 2 cos^{2} (x) - 7 cos (x) = 0

Solve by substitution

5 + 2 cos^{2} (x) - 7 cos (x) = 0

Let:

cos (x) = u

5 + 2 u^{2} - 7 u = 0

5 + 2 u^{2} - 7 u = 0 : u = \frac{5}{2}, u = 1

5 + 2 u^{2} - 7 u = 0

Write in the standard form

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

2 u^{2} - 7 u + 5 = 0

Solve with the quadratic formula

2 u^{2} - 7 u + 5 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = 5

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 5 = 3

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

Apply exponent rule:

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

n

is even

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

= 7^{2} - 4 \cdot 2 \cdot 5

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 7^{2} - 40

7^{2} = 49

= 49 - 40

Subtract the numbers:

49 - 40 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 7 ) \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 3}{2 \cdot 2}

u = \frac{- ( - 7 ) + 3}{2 \cdot 2} : \frac{5}{2}

\frac{- ( - 7 ) + 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 3}{2 \cdot 2}

Add the numbers:

7 + 3 = 10

= \frac{10}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{10}{4}

Cancel the common factor:

2

= \frac{5}{2}

u = \frac{- ( - 7 ) - 3}{2 \cdot 2} : 1

\frac{- ( - 7 ) - 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 3}{2 \cdot 2}

Subtract the numbers:

7 - 3 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{5}{2}, u = 1

Substitute back

u = cos (x)

cos (x) = \frac{5}{2}, cos (x) = 1

cos (x) = \frac{5}{2}, cos (x) = 1

cos (x) = \frac{5}{2} :

No Solution

cos (x) = \frac{5}{2}

- 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

csc^2(x)=2+csc(x)

csc^{2} (x) = 2 + csc (x)

tan(θ)=(5.4)/(1.1)

tan (θ) = \frac{5.4}{1.1}

6cos(2θ)+3=0

6 cos (2 θ) + 3 = 0

(sin(θ)+cos^2(θ))^2=2sin(θ)

(sin (θ) + cos^{2} (θ))^{2} = 2 sin (θ)

2/3 =sin(2θ)

\frac{2}{3} = sin (2 θ)

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)-7cos(x)+6=0 ?
The general solution for cos(2x)-7cos(x)+6=0 is x=2pin