What is the general solution for cos^2(x)=((7k-6))/4 ?

The general solution for cos^2(x)=((7k-6))/4 is

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cos^2(x)=((7k-6))/4

cos^{2} (x) = \frac{( 7 k - 6 )}{4}

x = arccos (\frac{7 k - 6}{2}) + 2 πn, x = - arccos (\frac{7 k - 6}{2}) + 2 πn, x = arccos (- \frac{7 k - 6}{2}) + 2 πn, x = - arccos (- \frac{7 k - 6}{2}) + 2 πn

Solution steps

cos^{2} (x) = \frac{( 7 k - 6 )}{4}

Solve by substitution

cos^{2} (x) = \frac{7 k - 6}{4}

Let:

cos (x) = u

u^{2} = \frac{7 k - 6}{4}

u^{2} = \frac{7 k - 6}{4} : u = \frac{7 k - 6}{2}, u = - \frac{7 k - 6}{2}

u^{2} = \frac{7 k - 6}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{7 k - 6}{4}, u = - \frac{7 k - 6}{4}

Simplify

\frac{7 k - 6}{4} : \frac{7 k - 6}{2}

\frac{7 k - 6}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7 k - 6}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{7 k - 6}{2}

Simplify

- \frac{7 k - 6}{4} : - \frac{7 k - 6}{2}

- \frac{7 k - 6}{4}

Simplify

\frac{7 k - 6}{4} : \frac{7 k - 6}{2}

\frac{7 k - 6}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7 k - 6}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{7 k - 6}{2}

= - \frac{7 k - 6}{2}

u = \frac{7 k - 6}{2}, u = - \frac{7 k - 6}{2}

Substitute back

u = cos (x)

cos (x) = \frac{7 k - 6}{2}, cos (x) = - \frac{7 k - 6}{2}

cos (x) = \frac{7 k - 6}{2}, cos (x) = - \frac{7 k - 6}{2}

cos (x) = \frac{7 k - 6}{2} : x = arccos (\frac{7 k - 6}{2}) + 2 πn, x = - arccos (\frac{7 k - 6}{2}) + 2 πn

cos (x) = \frac{7 k - 6}{2}

Apply trig inverse properties

cos (x) = \frac{7 k - 6}{2}

General solutions for

cos (x) = \frac{7 k - 6}{2}

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

x = arccos (\frac{7 k - 6}{2}) + 2 πn, x = - arccos (\frac{7 k - 6}{2}) + 2 πn

x = arccos (\frac{7 k - 6}{2}) + 2 πn, x = - arccos (\frac{7 k - 6}{2}) + 2 πn

cos (x) = - \frac{7 k - 6}{2} : x = arccos (- \frac{7 k - 6}{2}) + 2 πn, x = - arccos (- \frac{7 k - 6}{2}) + 2 πn

cos (x) = - \frac{7 k - 6}{2}

Apply trig inverse properties

cos (x) = - \frac{7 k - 6}{2}

General solutions for

cos (x) = - \frac{7 k - 6}{2}

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

x = arccos (- \frac{7 k - 6}{2}) + 2 πn, x = - arccos (- \frac{7 k - 6}{2}) + 2 πn

x = arccos (- \frac{7 k - 6}{2}) + 2 πn, x = - arccos (- \frac{7 k - 6}{2}) + 2 πn

Combine all the solutions

x = arccos (\frac{7 k - 6}{2}) + 2 πn, x = - arccos (\frac{7 k - 6}{2}) + 2 πn, x = arccos (- \frac{7 k - 6}{2}) + 2 πn, x = - arccos (- \frac{7 k - 6}{2}) + 2 πn

5 - 3 csc^{2} (x) = 1

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

tan (3 x) + cos (6 x) = 1

sec (θ) + tan (θ) = 2

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

What is the general solution for cos^2(x)=((7k-6))/4 ?
The general solution for cos^2(x)=((7k-6))/4 is