What is the general solution for 8cos(2x)=8cos(x)-9 ?

The general solution for 8cos(2x)=8cos(x)-9 is x=1.31811…+2pin,x=2pi-1.31811…+2pin

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Popular Trigonometry >

8cos(2x)=8cos(x)-9

Solution

8 cos (2 x) = 8 cos (x) - 9

Solution

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

Degrees

x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 cos (2 x) = 8 cos (x) - 9

Subtract

8 cos (x) - 9

from both sides

8 cos (2 x) - 8 cos (x) + 9 = 0

Rewrite using trig identities

9 + 8 cos (2 x) - 8 cos (x)

Use the Double Angle identity:

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

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

Simplify

9 + 8 (2 cos^{2} (x) - 1) - 8 cos (x) : 16 cos^{2} (x) - 8 cos (x) + 1

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

Expand

8 (2 cos^{2} (x) - 1) : 16 cos^{2} (x) - 8

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

Apply the distributive law:

a (b - c) = ab - a c

a = 8, b = 2 cos^{2} (x), c = 1

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

Simplify

8 \cdot 2 cos^{2} (x) - 8 \cdot 1 : 16 cos^{2} (x) - 8

8 \cdot 2 cos^{2} (x) - 8 \cdot 1

Multiply the numbers:

8 \cdot 2 = 16

= 16 cos^{2} (x) - 8 \cdot 1

Multiply the numbers:

8 \cdot 1 = 8

= 16 cos^{2} (x) - 8

= 16 cos^{2} (x) - 8

= 9 + 16 cos^{2} (x) - 8 - 8 cos (x)

Simplify

9 + 16 cos^{2} (x) - 8 - 8 cos (x) : 16 cos^{2} (x) - 8 cos (x) + 1

9 + 16 cos^{2} (x) - 8 - 8 cos (x)

Group like terms

= 16 cos^{2} (x) - 8 cos (x) + 9 - 8

Add/Subtract the numbers:

9 - 8 = 1

= 16 cos^{2} (x) - 8 cos (x) + 1

= 16 cos^{2} (x) - 8 cos (x) + 1

= 16 cos^{2} (x) - 8 cos (x) + 1

1 + 16 cos^{2} (x) - 8 cos (x) = 0

Solve by substitution

1 + 16 cos^{2} (x) - 8 cos (x) = 0

Let:

cos (x) = u

1 + 16 u^{2} - 8 u = 0

1 + 16 u^{2} - 8 u = 0 : u = \frac{1}{4}

1 + 16 u^{2} - 8 u = 0

Write in the standard form

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

16 u^{2} - 8 u + 1 = 0

Solve with the quadratic formula

16 u^{2} - 8 u + 1 = 0

Quadratic Equation Formula:

For

a = 16, b = - 8, c = 1

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 16 \cdot 1}{2 \cdot 16}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 16 \cdot 1}{2 \cdot 16}

(- 8)^{2} - 4 \cdot 16 \cdot 1 = 0

(- 8)^{2} - 4 \cdot 16 \cdot 1

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 16 \cdot 1

Multiply the numbers:

4 \cdot 16 \cdot 1 = 64

= 8^{2} - 64

8^{2} = 64

= 64 - 64

Subtract the numbers:

64 - 64 = 0

= 0

u_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 16}

u = \frac{- ( - 8 )}{2 \cdot 16}

\frac{- ( - 8 )}{2 \cdot 16} = \frac{1}{4}

\frac{- ( - 8 )}{2 \cdot 16}

Apply rule

- (- a) = a

= \frac{8}{2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{8}{32}

Cancel the common factor:

8

= \frac{1}{4}

u = \frac{1}{4}

The solution to the quadratic equation is:

u = \frac{1}{4}

Substitute back

u = cos (x)

cos (x) = \frac{1}{4}

cos (x) = \frac{1}{4}

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

cos (x) = \frac{1}{4}

Apply trig inverse properties

cos (x) = \frac{1}{4}

General solutions for

cos (x) = \frac{1}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 8cos(2x)=8cos(x)-9 ?
The general solution for 8cos(2x)=8cos(x)-9 is x=1.31811…+2pin,x=2pi-1.31811…+2pin