What is the general solution for cos(2x)-5cos(x)+3=0 ?

The general solution for cos(2x)-5cos(x)+3=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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cos(2x)-5cos(x)+3=0

Solution

cos (2 x) - 5 cos (x) + 3 = 0

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) - 5 cos (x) + 3 = 0

Rewrite using trig identities

3 + cos (2 x) - 5 cos (x)

Use the Double Angle identity:

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

= 3 + 2 cos^{2} (x) - 1 - 5 cos (x)

Simplify

3 + 2 cos^{2} (x) - 1 - 5 cos (x) : 2 cos^{2} (x) - 5 cos (x) + 2

3 + 2 cos^{2} (x) - 1 - 5 cos (x)

Group like terms

= 2 cos^{2} (x) - 5 cos (x) + 3 - 1

Add/Subtract the numbers:

3 - 1 = 2

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

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

\frac{- ( - 5 ) + 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 3}{2 \cdot 2}

Add the numbers:

5 + 3 = 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2

u = \frac{- ( - 5 ) - 3}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 5 ) - 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 3}{2 \cdot 2}

Subtract the numbers:

5 - 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

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

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

9sin^2(x)+18sin(x)+9=0

9 sin^{2} (x) + 18 sin (x) + 9 = 0

-12sin^2(θ)+7=4

- 12 sin^{2} (θ) + 7 = 4

4cos(θ)+5=0

4 cos (θ) + 5 = 0

cos(θ)= 60/78

cos (θ) = \frac{60}{78}

18sin^2(x)=9

18 sin^{2} (x) = 9

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)-5cos(x)+3=0 ?
The general solution for cos(2x)-5cos(x)+3=0 is x= pi/3+2pin,x=(5pi)/3+2pin