What is the general solution for cos(2x)=cos(x)-cos(30)[180] ?

The general solution for cos(2x)=cos(x)-cos(30)[180] is No Solution for x\in\mathbb{R}

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Popular >

cos(2x)=cos(x)-cos(30)[180]

Solution

cos (2 x) = cos (x) - cos (3 0^{\circ}) [180]

Solution

No Solution for x \in R

Solution steps

cos (2 x) = cos (x) - cos (3 0^{\circ}) [180]

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

cos (2 x) = cos (x) - \frac{3}{2} [180]

Subtract

cos (x) - \frac{3}{2} [180]

from both sides

cos (2 x) - cos (x) + 903 = 0

Rewrite using trig identities

cos (2 x) - cos (x) + 903

Use the Double Angle identity:

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

= 2 cos^{2} (x) - 1 - cos (x) + 903

- 1 - cos (x) + 2 cos^{2} (x) + 903 = 0

Solve by substitution

- 1 - cos (x) + 2 cos^{2} (x) + 903 = 0

Let:

cos (x) = u

- 1 - u + 2 u^{2} + 903 = 0

- 1 - u + 2 u^{2} + 903 = 0 : u = \frac{1}{4} + i \frac{3 - 1 + 80 3}{4}, u = \frac{1}{4} - i \frac{3 - 1 + 80 3}{4}

- 1 - u + 2 u^{2} + 903 = 0

Write in the standard form

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

2 u^{2} - u - 1 + 903 = 0

Solve with the quadratic formula

2 u^{2} - u - 1 + 903 = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = - 1 + 903

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 1 + 90 3 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 1 + 90 3 )}{2 \cdot 2}

Simplify

(- 1)^{2} - 4 \cdot 2 (- 1 + 903) : 3 i 803 - 1

(- 1)^{2} - 4 \cdot 2 (- 1 + 903)

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 (- 1 + 903) = 8 (- 1 + 903)

4 \cdot 2 (- 1 + 903)

Multiply the numbers:

4 \cdot 2 = 8

= 8 (903 - 1)

= 1 - 8 (903 - 1)

Apply imaginary number rule:

- a = i a

= i 8 (903 - 1) - 1

- 1 + 8 (- 1 + 903) = 3 803 - 1

- 1 + 8 (- 1 + 903)

Expand

- 1 + 8 (- 1 + 903) : 7203 - 9

- 1 + 8 (- 1 + 903)

Expand

8 (- 1 + 903) : - 8 + 7203

8 (- 1 + 903)

Apply the distributive law:

a (b + c) = ab + a c

a = 8, b = - 1, c = 903

= 8 (- 1) + 8 \cdot 903

Apply minus-plus rules

+ (- a) = - a

= - 8 \cdot 1 + 8 \cdot 903

Simplify

- 8 \cdot 1 + 8 \cdot 903 : - 8 + 7203

- 8 \cdot 1 + 8 \cdot 903

Multiply the numbers:

8 \cdot 1 = 8

= - 8 + 8 \cdot 903

Multiply the numbers:

8 \cdot 90 = 720

= - 8 + 7203

= - 8 + 7203

= - 1 - 8 + 7203

Subtract the numbers:

- 1 - 8 = - 9

= 7203 - 9

= 7203 - 9

Factor

7203 - 9 : 9 (803 - 1)

7203 - 9

Rewrite as

= 9 \cdot 803 - 9 \cdot 1

Factor out common term

9

= 9 (803 - 1)

= 9 (803 - 1)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 9 803 - 1

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= 3 803 - 1

= 3 i 803 - 1

u_{1, 2} = \frac{- ( - 1 ) \pm 3 i 80 3 - 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3 i 80 3 - 1}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 3 i 80 3 - 1}{2 \cdot 2}

u = \frac{- ( - 1 ) + 3 i 80 3 - 1}{2 \cdot 2} : \frac{1}{4} + i \frac{3 - 1 + 80 3}{4}

\frac{- ( - 1 ) + 3 i 80 3 - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 3 i 80 3 - 1}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 3 i 80 3 - 1}{4}

Rewrite

\frac{1 + 3 i 80 3 - 1}{4}

in standard complex form:

\frac{1}{4} + \frac{3 80 3 - 1}{4} i

\frac{1 + 3 i 80 3 - 1}{4}

Apply the fraction rule:

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

\frac{1 + 3 i 80 3 - 1}{4} = \frac{1}{4} + \frac{3 i 80 3 - 1}{4}

= \frac{1}{4} + \frac{3 i 80 3 - 1}{4}

= \frac{1}{4} + \frac{3 80 3 - 1}{4} i

u = \frac{- ( - 1 ) - 3 i 80 3 - 1}{2 \cdot 2} : \frac{1}{4} - i \frac{3 - 1 + 80 3}{4}

\frac{- ( - 1 ) - 3 i 80 3 - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 3 i 80 3 - 1}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 3 i 80 3 - 1}{4}

Rewrite

\frac{1 - 3 i 80 3 - 1}{4}

in standard complex form:

\frac{1}{4} - \frac{3 80 3 - 1}{4} i

\frac{1 - 3 i 80 3 - 1}{4}

Apply the fraction rule:

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

\frac{1 - 3 i 80 3 - 1}{4} = \frac{1}{4} - \frac{3 i 80 3 - 1}{4}

= \frac{1}{4} - \frac{3 i 80 3 - 1}{4}

= \frac{1}{4} - \frac{3 80 3 - 1}{4} i

The solutions to the quadratic equation are:

u = \frac{1}{4} + i \frac{3 - 1 + 80 3}{4}, u = \frac{1}{4} - i \frac{3 - 1 + 80 3}{4}

Substitute back

u = cos (x)

cos (x) = \frac{1}{4} + i \frac{3 - 1 + 80 3}{4}, cos (x) = \frac{1}{4} - i \frac{3 - 1 + 80 3}{4}

cos (x) = \frac{1}{4} + i \frac{3 - 1 + 80 3}{4}, cos (x) = \frac{1}{4} - i \frac{3 - 1 + 80 3}{4}

cos (x) = \frac{1}{4} + i \frac{3 - 1 + 80 3}{4} :

No Solution

cos (x) = \frac{1}{4} + i \frac{3 - 1 + 80 3}{4}

No Solution

cos (x) = \frac{1}{4} - i \frac{3 - 1 + 80 3}{4} :

No Solution

cos (x) = \frac{1}{4} - i \frac{3 - 1 + 80 3}{4}

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)=cos(x)-cos(30)[180] ?
The general solution for cos(2x)=cos(x)-cos(30)[180] is No Solution for x\in\mathbb{R}