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

The general solution for cos^2(x)=((8k-9))/9 is

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cos^2(x)=((8k-9))/9

cos^{2} (x) = \frac{( 8 k - 9 )}{9}

x = arccos (\frac{8 k - 9}{3}) + 2 πn, x = - arccos (\frac{8 k - 9}{3}) + 2 πn, x = arccos (- \frac{8 k - 9}{3}) + 2 πn, x = - arccos (- \frac{8 k - 9}{3}) + 2 πn

Solution steps

cos^{2} (x) = \frac{( 8 k - 9 )}{9}

Solve by substitution

cos^{2} (x) = \frac{8 k - 9}{9}

Let:

cos (x) = u

u^{2} = \frac{8 k - 9}{9}

u^{2} = \frac{8 k - 9}{9} : u = \frac{8 k - 9}{3}, u = - \frac{8 k - 9}{3}

u^{2} = \frac{8 k - 9}{9}

For

x^{2} = f (a)

the solutions are

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

u = \frac{8 k - 9}{9}, u = - \frac{8 k - 9}{9}

Simplify

\frac{8 k - 9}{9} : \frac{8 k - 9}{3}

\frac{8 k - 9}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{8 k - 9}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{8 k - 9}{3}

Simplify

- \frac{8 k - 9}{9} : - \frac{8 k - 9}{3}

- \frac{8 k - 9}{9}

Simplify

\frac{8 k - 9}{9} : \frac{8 k - 9}{3}

\frac{8 k - 9}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{8 k - 9}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{8 k - 9}{3}

= - \frac{8 k - 9}{3}

u = \frac{8 k - 9}{3}, u = - \frac{8 k - 9}{3}

Substitute back

u = cos (x)

cos (x) = \frac{8 k - 9}{3}, cos (x) = - \frac{8 k - 9}{3}

cos (x) = \frac{8 k - 9}{3}, cos (x) = - \frac{8 k - 9}{3}

cos (x) = \frac{8 k - 9}{3} : x = arccos (\frac{8 k - 9}{3}) + 2 πn, x = - arccos (\frac{8 k - 9}{3}) + 2 πn

cos (x) = \frac{8 k - 9}{3}

Apply trig inverse properties

cos (x) = \frac{8 k - 9}{3}

General solutions for

cos (x) = \frac{8 k - 9}{3}

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

x = arccos (\frac{8 k - 9}{3}) + 2 πn, x = - arccos (\frac{8 k - 9}{3}) + 2 πn

x = arccos (\frac{8 k - 9}{3}) + 2 πn, x = - arccos (\frac{8 k - 9}{3}) + 2 πn

cos (x) = - \frac{8 k - 9}{3} : x = arccos (- \frac{8 k - 9}{3}) + 2 πn, x = - arccos (- \frac{8 k - 9}{3}) + 2 πn

cos (x) = - \frac{8 k - 9}{3}

Apply trig inverse properties

cos (x) = - \frac{8 k - 9}{3}

General solutions for

cos (x) = - \frac{8 k - 9}{3}

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

x = arccos (- \frac{8 k - 9}{3}) + 2 πn, x = - arccos (- \frac{8 k - 9}{3}) + 2 πn

x = arccos (- \frac{8 k - 9}{3}) + 2 πn, x = - arccos (- \frac{8 k - 9}{3}) + 2 πn

Combine all the solutions

x = arccos (\frac{8 k - 9}{3}) + 2 πn, x = - arccos (\frac{8 k - 9}{3}) + 2 πn, x = arccos (- \frac{8 k - 9}{3}) + 2 πn, x = - arccos (- \frac{8 k - 9}{3}) + 2 πn

sin (x) + cot (x) cos (x) = 3

4 sin^{4} (x) + 3 sin^{2} (x) - 1 = 0

3 tan (θ) - 3 = 0

cot (θ) = \frac{1}{cot ( θ )}

(tan (x))^{5} - 9 tan (x) = 0

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