What is the general solution for cos^2(x)=((4k-3)/9) ?

The general solution for cos^2(x)=((4k-3)/9) is

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cos^2(x)=((4k-3)/9)

cos^{2} (x) = (\frac{4 k - 3}{9})

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

Solution steps

cos^{2} (x) = (\frac{4 k - 3}{9})

Solve by substitution

cos^{2} (x) = \frac{4 k - 3}{9}

Let:

cos (x) = u

u^{2} = \frac{4 k - 3}{9}

u^{2} = \frac{4 k - 3}{9} : u = \frac{4 k - 3}{3}, u = - \frac{4 k - 3}{3}

u^{2} = \frac{4 k - 3}{9}

For

x^{2} = f (a)

the solutions are

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

u = \frac{4 k - 3}{9}, u = - \frac{4 k - 3}{9}

Simplify

\frac{4 k - 3}{9} : \frac{4 k - 3}{3}

\frac{4 k - 3}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4 k - 3}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Simplify

- \frac{4 k - 3}{9} : - \frac{4 k - 3}{3}

- \frac{4 k - 3}{9}

Simplify

\frac{4 k - 3}{9} : \frac{4 k - 3}{3}

\frac{4 k - 3}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4 k - 3}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

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

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

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

9 = - 9 \cdot sin (2 x)

2 sin (4 t) = 2 sin (4 t) cos (4 t)

a^{1} sin (x) = 43.1253

52800 = \frac{( 1500 ) ^{2} sin ( x )}{32.2}

7 cos (3 x) = 2, 0 \leq x < 2 π

What is the general solution for cos^2(x)=((4k-3)/9) ?
The general solution for cos^2(x)=((4k-3)/9) is