What is the general solution for cos(2x)+sqrt(3)cos(x)=2 ?

The general solution for cos(2x)+sqrt(3)cos(x)=2 is x= pi/6+2pin,x=(11pi)/6+2pin

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cos(2x)+sqrt(3)cos(x)=2

Solution

cos (2 x) + 3 cos (x) = 2

Solution

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) + 3 cos (x) = 2

Subtract

2

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 2 - 1 = - 3

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

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

- 3 + 2 cos^{2} (x) + cos (x) 3 = 0

Solve by substitution

- 3 + 2 cos^{2} (x) + cos (x) 3 = 0

Let:

cos (x) = u

- 3 + 2 u^{2} + u 3 = 0

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

- 3 + 2 u^{2} + u 3 = 0

Write in the standard form

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

2 u^{2} + 3 u - 3 = 0

Solve with the quadratic formula

2 u^{2} + 3 u - 3 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = - 3

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

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

(3)^{2} - 4 \cdot 2 (- 3) = 33

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

Apply rule

- (- a) = a

= (3)^{2} + 4 \cdot 2 \cdot 3

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

4 \cdot 2 \cdot 3 = 24

4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 24

= 3 + 24

Add the numbers:

3 + 24 = 27

= 27

Prime factorization of

27 : 3^{3}

27

27

divides by

327 = 9 \cdot 3

= 3 \cdot 9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3 \cdot 3

3

is a prime number, therefore no further factorization is possible

= 3 \cdot 3 \cdot 3

= 3^{3}

= 3^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 3^{2} \cdot 3

Apply radical rule:

= 3 3^{2}

Apply radical rule:

3^{2} = 3

= 33

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

Separate the solutions

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

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

\frac{- 3 + 3 3}{2 \cdot 2}

Add similar elements:

- 3 + 33 = 23

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Cancel the common factor:

2

= \frac{3}{2}

u = \frac{- 3 - 3 3}{2 \cdot 2} : - 3

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

Add similar elements:

- 3 - 33 = - 43

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4 3}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4 3}{4}

Divide the numbers:

\frac{4}{4} = 1

= - 3

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{3}{2} : x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

General solutions for

cos (x) = \frac{3}{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}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

cos (x) = - 3 :

No Solution

cos (x) = - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)+sqrt(3)cos(x)=2 ?
The general solution for cos(2x)+sqrt(3)cos(x)=2 is x= pi/6+2pin,x=(11pi)/6+2pin