What is the general solution for cos^2(x)=((4k-5))/(10) ?

The general solution for cos^2(x)=((4k-5))/(10) is

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cos^2(x)=((4k-5))/(10)

cos^{2} (x) = \frac{( 4 k - 5 )}{10}

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn, x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

Solution steps

cos^{2} (x) = \frac{( 4 k - 5 )}{10}

Solve by substitution

cos^{2} (x) = \frac{4 k - 5}{10}

Let:

cos (x) = u

u^{2} = \frac{4 k - 5}{10}

u^{2} = \frac{4 k - 5}{10} : u = \frac{4 k - 5}{10}, u = - \frac{4 k - 5}{10}

u^{2} = \frac{4 k - 5}{10}

For

x^{2} = f (a)

the solutions are

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

u = \frac{4 k - 5}{10}, u = - \frac{4 k - 5}{10}

Substitute back

u = cos (x)

cos (x) = \frac{4 k - 5}{10}, cos (x) = - \frac{4 k - 5}{10}

cos (x) = \frac{4 k - 5}{10}, cos (x) = - \frac{4 k - 5}{10}

cos (x) = \frac{4 k - 5}{10} : x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

cos (x) = \frac{4 k - 5}{10}

Apply trig inverse properties

cos (x) = \frac{4 k - 5}{10}

General solutions for

cos (x) = \frac{4 k - 5}{10}

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

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

cos (x) = - \frac{4 k - 5}{10} : x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

cos (x) = - \frac{4 k - 5}{10}

Apply trig inverse properties

cos (x) = - \frac{4 k - 5}{10}

General solutions for

cos (x) = - \frac{4 k - 5}{10}

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

x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

Combine all the solutions

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn, x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

csc (A) = 3

2.08 \cdot cos (θ) - 9.78 \cdot sin (θ) = 0.836

cos (x) = \frac{0.95293}{1.36156}

θ, W = F \cdot d \cdot cos (θ)

cos (\frac{π}{6} + x) + sin (\frac{π}{3} + x) = 0

What is the general solution for cos^2(x)=((4k-5))/(10) ?
The general solution for cos^2(x)=((4k-5))/(10) is