What is the general solution for 2sin^2(4x)-2cos(4x)=1,o<= x<= 360 ?

Q: What is the general solution for 2sin^2(4x)-2cos(4x)=1,o<= x<= 360 ?

The general solution for 2sin^2(4x)-2cos(4x)=1,o<= x<= 360 is No Solution for x\in\mathbb{R}

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2sin^2(4x)-2cos(4x)=1,o<= x<= 360

Solution

2 sin^{2} (4 x) - 2 cos (4 x) = 1, o^{\circ} \leq x \leq 36 0^{\circ}

Solution

No Solution for x \in R

Solution steps

2 sin^{2} (4 x) - 2 cos (4 x) = 1, o^{\circ} \leq x \leq 36 0^{\circ}

Subtract

1

from both sides

2 sin^{2} (4 x) - 2 cos (4 x) - 1 = 0

Rewrite using trig identities

- 1 - 2 cos (4 x) + 2 sin^{2} (4 x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (4 x)

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

1 - 2 cos (4 x) - 2 cos^{2} (4 x) = 0

Solve by substitution

1 - 2 cos (4 x) - 2 cos^{2} (4 x) = 0

Let:

cos (4 x) = u

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

1 - 2 u - 2 u^{2} = 0 : u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 2)^{2} - 4 (- 2) \cdot 1 = 23

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

Add the numbers:

4 + 8 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

n ab = n a n b

= 3 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 23

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

Separate the solutions

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

u = \frac{- ( - 2 ) + 2 3}{2 ( - 2 )} : - \frac{1 + 3}{2}

\frac{- ( - 2 ) + 2 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

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

Apply the fraction rule:

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

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

Cancel

\frac{2 + 2 3}{4} : \frac{1 + 3}{2}

\frac{2 + 2 3}{4}

Factor

2 + 23 : 2 (1 + 3)

2 + 23

Rewrite as

= 2 \cdot 1 + 23

Factor out common term

2

= 2 (1 + 3)

= \frac{2 ( 1 + 3 )}{4}

Cancel the common factor:

2

= \frac{1 + 3}{2}

= - \frac{1 + 3}{2}

u = \frac{- ( - 2 ) - 2 3}{2 ( - 2 )} : \frac{3 - 1}{2}

\frac{- ( - 2 ) - 2 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

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

Apply the fraction rule:

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

2 - 23 = - (23 - 2)

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

Factor

23 - 2 : 2 (3 - 1)

23 - 2

Rewrite as

= 23 - 2 \cdot 1

Factor out common term

2

= 2 (3 - 1)

= \frac{2 ( 3 - 1 )}{4}

Cancel the common factor:

2

= \frac{3 - 1}{2}

The solutions to the quadratic equation are:

u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

Substitute back

u = cos (4 x)

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

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

cos (4 x) = - \frac{1 + 3}{2}, o^{\circ} \leq x \leq 36 0^{\circ} :

No Solution

cos (4 x) = - \frac{1 + 3}{2}, o^{\circ} \leq x \leq 36 0^{\circ}

- 1 \leq cos (x) \leq 1

No Solution

cos (4 x) = \frac{3 - 1}{2}, o^{\circ} \leq x \leq 36 0^{\circ} :

No Solution

cos (4 x) = \frac{3 - 1}{2}, o^{\circ} \leq x \leq 36 0^{\circ}

Apply trig inverse properties

cos (4 x) = \frac{3 - 1}{2}

General solutions for

cos (4 x) = \frac{3 - 1}{2}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

4 x = arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n, 4 x = 36 0^{\circ} - arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n

4 x = arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n, 4 x = 36 0^{\circ} - arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n

Solve

4 x = arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n : x = \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

4 x = arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n

Divide both sides by

4

4 x = arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n

Divide both sides by

4

\frac{4 x}{4} = \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{36 0 ^{\circ} n}{4}

Simplify

x = \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

x = \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

Solve

4 x = 36 0^{\circ} - arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n : x = 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

4 x = 36 0^{\circ} - arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n

Divide both sides by

4

4 x = 36 0^{\circ} - arccos (\frac{3 - 1}{2}) + 36 0^{\circ} n

Divide both sides by

4

\frac{4 x}{4} = 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{36 0 ^{\circ} n}{4}

Simplify

\frac{4 x}{4} = 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{36 0 ^{\circ} n}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{36 0 ^{\circ} n}{4} : 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{36 0 ^{\circ} n}{4}

Cancel

9 0^{\circ} : 9 0^{\circ}

9 0^{\circ}

Cancel the common factor:

2

= 9 0^{\circ}

= 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{36 0 ^{\circ} n}{4}

Cancel

\frac{36 0 ^{\circ} n}{4} : \frac{18 0 ^{\circ} n}{2}

\frac{36 0 ^{\circ} n}{4}

Cancel the common factor:

2

= \frac{18 0 ^{\circ} n}{2}

= 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

x = 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

x = 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

x = 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

x = \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}, x = 9 0^{\circ} - \frac{arccos ( \frac{3 - 1}{2} )}{4} + \frac{18 0 ^{\circ} n}{2}

Solutions for the range

o^{\circ} \leq x \leq 36 0^{\circ}

No Solution

Combine all the solutions

No Solution for x \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(4x)-2cos(4x)=1,o<= x<= 360 ?
The general solution for 2sin^2(4x)-2cos(4x)=1,o<= x<= 360 is No Solution for x\in\mathbb{R}