What is the general solution for cos(2x)-7cos(x)-3=0 ?

The general solution for cos(2x)-7cos(x)-3=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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cos(2x)-7cos(x)-3=0

Solution

cos (2 x) - 7 cos (x) - 3 = 0

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) - 7 cos (x) - 3 = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

= - 3 + 2 cos^{2} (x) - 1 - 7 cos (x)

Simplify

- 3 + 2 cos^{2} (x) - 1 - 7 cos (x) : 2 cos^{2} (x) - 7 cos (x) - 4

- 3 + 2 cos^{2} (x) - 1 - 7 cos (x)

Group like terms

= 2 cos^{2} (x) - 7 cos (x) - 3 - 1

Subtract the numbers:

- 3 - 1 = - 4

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 7, c = - 4

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

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

(- 7)^{2} - 4 \cdot 2 (- 4) = 9

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 7^{2} + 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

Add the numbers:

49 + 32 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 7 ) \pm 9}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 9}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 9}{2 \cdot 2}

u = \frac{- ( - 7 ) + 9}{2 \cdot 2} : 4

\frac{- ( - 7 ) + 9}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 9}{2 \cdot 2}

Add the numbers:

7 + 9 = 16

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{16}{4}

Divide the numbers:

\frac{16}{4} = 4

= 4

u = \frac{- ( - 7 ) - 9}{2 \cdot 2} : - \frac{1}{2}

\frac{- ( - 7 ) - 9}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 9}{2 \cdot 2}

Subtract the numbers:

7 - 9 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = 4, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = 4, cos (x) = - \frac{1}{2}

cos (x) = 4, cos (x) = - \frac{1}{2}

cos (x) = 4 :

No Solution

cos (x) = 4

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

tan^2(45-x/3)=0.58

tan^{2} (4 5^{\circ} - \frac{x}{3}) = 0.58

-20csc(x)=40

- 20 csc (x) = 40

0=4|sin(6θ)|

0 = 4 ∣ sin (6 θ) ∣

5sin^2(θ)-11sin(θ)+6=0

5 sin^{2} (θ) - 11 sin (θ) + 6 = 0

cos(x)= 11/14 ,sin(x)

cos (x) = \frac{11}{14}, sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)-7cos(x)-3=0 ?
The general solution for cos(2x)-7cos(x)-3=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin