What is the general solution for cos(4x)-7cos(2x)=8 ?

The general solution for cos(4x)-7cos(2x)=8 is x= pi/2+pin

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Popular >

cos(4x)-7cos(2x)=8

Solution

cos (4 x) - 7 cos (2 x) = 8

Solution

x = \frac{π}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cos (4 x) - 7 cos (2 x) = 8

Subtract

8

from both sides

cos (4 x) - 7 cos (2 x) - 8 = 0

Rewrite using trig identities

- 8 + cos (4 x) - 7 cos (2 x)

cos (4 x) = 2 cos^{2} (2 x) - 1

cos (4 x)

Rewrite as

= cos (2 \cdot 2 x)

Use the Double Angle identity:

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

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

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

= - 8 + 2 cos^{2} (2 x) - 1 - 7 cos (2 x)

Simplify

- 8 + 2 cos^{2} (2 x) - 1 - 7 cos (2 x) : 2 cos^{2} (2 x) - 7 cos (2 x) - 9

- 8 + 2 cos^{2} (2 x) - 1 - 7 cos (2 x)

Group like terms

= 2 cos^{2} (2 x) - 7 cos (2 x) - 8 - 1

Subtract the numbers:

- 8 - 1 = - 9

= 2 cos^{2} (2 x) - 7 cos (2 x) - 9

= 2 cos^{2} (2 x) - 7 cos (2 x) - 9

- 9 + 2 cos^{2} (2 x) - 7 cos (2 x) = 0

Solve by substitution

- 9 + 2 cos^{2} (2 x) - 7 cos (2 x) = 0

Let:

cos (2 x) = u

- 9 + 2 u^{2} - 7 u = 0

- 9 + 2 u^{2} - 7 u = 0 : u = \frac{9}{2}, u = - 1

- 9 + 2 u^{2} - 7 u = 0

Write in the standard form

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

2 u^{2} - 7 u - 9 = 0

Solve with the quadratic formula

2 u^{2} - 7 u - 9 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = - 9

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 9) = 11

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 7^{2} + 4 \cdot 2 \cdot 9

Multiply the numbers:

4 \cdot 2 \cdot 9 = 72

= 7^{2} + 72

7^{2} = 49

= 49 + 72

Add the numbers:

49 + 72 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

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

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 11}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 11}{2 \cdot 2}

u = \frac{- ( - 7 ) + 11}{2 \cdot 2} : \frac{9}{2}

\frac{- ( - 7 ) + 11}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 11}{2 \cdot 2}

Add the numbers:

7 + 11 = 18

= \frac{18}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{18}{4}

Cancel the common factor:

2

= \frac{9}{2}

u = \frac{- ( - 7 ) - 11}{2 \cdot 2} : - 1

\frac{- ( - 7 ) - 11}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 11}{2 \cdot 2}

Subtract the numbers:

7 - 11 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{9}{2}, u = - 1

Substitute back

u = cos (2 x)

cos (2 x) = \frac{9}{2}, cos (2 x) = - 1

cos (2 x) = \frac{9}{2}, cos (2 x) = - 1

cos (2 x) = \frac{9}{2} :

No Solution

cos (2 x) = \frac{9}{2}

- 1 \leq cos (x) \leq 1

No Solution

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

cos (2 x) = - 1

General solutions for

cos (2 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}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

Combine all the solutions

x = \frac{π}{2} + πn

Popular Examples

6tan(B)+sqrt(14)=0

6 tan (B) + 14 = 0

-4tan^2(θ)-4tan(θ)-9=8tan(θ)-4

- 4 tan^{2} (θ) - 4 tan (θ) - 9 = 8 tan (θ) - 4

cot(4x)=-(sqrt(3))/3

cot (4 x) = - \frac{3}{3}

sin(x)=(3.9)/(8.1)

sin (x) = \frac{3.9}{8.1}

5sin(x)=-2cos^2(x)+4

5 sin (x) = - 2 cos^{2} (x) + 4

Frequently Asked Questions (FAQ)

What is the general solution for cos(4x)-7cos(2x)=8 ?
The general solution for cos(4x)-7cos(2x)=8 is x= pi/2+pin