What is the general solution for cos^2(x)=((4k-3))/8 ?

The general solution for cos^2(x)=((4k-3))/8 is

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cos^2(x)=((4k-3))/8

Solution

cos^{2} (x) = \frac{( 4 k - 3 )}{8}

Solution

x = arccos (\frac{2 4 k - 3}{4}) + 2 πn, x = - arccos (\frac{2 4 k - 3}{4}) + 2 πn, x = arccos (- \frac{2 4 k - 3}{4}) + 2 πn, x = - arccos (- \frac{2 4 k - 3}{4}) + 2 πn

Solution steps

cos^{2} (x) = \frac{( 4 k - 3 )}{8}

Solve by substitution

cos^{2} (x) = \frac{4 k - 3}{8}

Let:

cos (x) = u

u^{2} = \frac{4 k - 3}{8}

u^{2} = \frac{4 k - 3}{8} : u = \frac{2 4 k - 3}{4}, u = - \frac{2 4 k - 3}{4}

u^{2} = \frac{4 k - 3}{8}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{4 k - 3}{8}, u = - \frac{4 k - 3}{8}

Simplify

\frac{4 k - 3}{8} : \frac{2 4 k - 3}{4}

\frac{4 k - 3}{8}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4 k - 3}{8}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Rationalize

\frac{4 k - 3}{2 2} : \frac{2 4 k - 3}{4}

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

Multiply by the conjugate

\frac{2}{2}

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

222 = 4

222

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

Add similar elements:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

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

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2^{1 + 1}

Refine

= 4

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

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

Simplify

- \frac{4 k - 3}{8} : - \frac{2 4 k - 3}{4}

- \frac{4 k - 3}{8}

Simplify

\frac{4 k - 3}{8} : \frac{4 k - 3}{2 2}

\frac{4 k - 3}{8}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4 k - 3}{8}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

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

Rationalize

- \frac{4 k - 3}{2 2} : - \frac{2 4 k - 3}{4}

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

Multiply by the conjugate

\frac{2}{2}

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

222 = 4

222

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

Add similar elements:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

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

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2^{1 + 1}

Refine

= 4

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

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

u = \frac{2 4 k - 3}{4}, u = - \frac{2 4 k - 3}{4}

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arccos (\frac{2 4 k - 3}{4}) + 2 πn, x = - arccos (\frac{2 4 k - 3}{4}) + 2 πn

x = arccos (\frac{2 4 k - 3}{4}) + 2 πn, x = - arccos (\frac{2 4 k - 3}{4}) + 2 πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arccos (- \frac{2 4 k - 3}{4}) + 2 πn, x = - arccos (- \frac{2 4 k - 3}{4}) + 2 πn

x = arccos (- \frac{2 4 k - 3}{4}) + 2 πn, x = - arccos (- \frac{2 4 k - 3}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{2 4 k - 3}{4}) + 2 πn, x = - arccos (\frac{2 4 k - 3}{4}) + 2 πn, x = arccos (- \frac{2 4 k - 3}{4}) + 2 πn, x = - arccos (- \frac{2 4 k - 3}{4}) + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)=((4k-3))/8 ?
The general solution for cos^2(x)=((4k-3))/8 is