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

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

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

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

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

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

For

x^{2} = f (a)

the solutions are

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

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

Simplify

\frac{2 k - 8}{9} : \frac{2 k - 8}{3}

\frac{2 k - 8}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{2 k - 8}{3}

Simplify

- \frac{2 k - 8}{9} : - \frac{2 k - 8}{3}

- \frac{2 k - 8}{9}

Simplify

\frac{2 k - 8}{9} : \frac{2 k - 8}{3}

\frac{2 k - 8}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{2 k - 8}{3}

= - \frac{2 k - 8}{3}

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

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} = sec (x)

sin^{2} (x) = 2 + 2 cos (x), 0 \leq x \leq 2 π

cos (5 x + 3) = \frac{1}{2}

csc (θ) = 3

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

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