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

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

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

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

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

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

For

x^{2} = f (a)

the solutions are

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

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

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

(\frac{1 + tanh ( x )}{1 - tanh ( x )})^{2} = 2

cos (2 x) - cos (- x) = 0

cos (3 x) = sin (x)

4 tan^{2} (θ) = 12, 0 \leq θ < 2 π

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

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