What is the general solution for cos^2(x)=((4k-3))/8 ?

The general solution for cos^2(x)=((4k-3))/8 is

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cos^2(x)=((4k-3))/8

cos^{2} (x) = \frac{( 4 k - 3 )}{8}

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

Solution steps

cos^{2} (x) = \frac{( 4 k - 3 )}{8}

Solve by substitution

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

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

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

Combine all the solutions

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

3 cos^{2} (x) - 8 cos (x) + 4 = 0

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

49.55 \cdot 1 - sin^{2} (θ) - 30 sin (θ) = 1.225

cos (2 x) + sin (x) - 1 = 0

x, e^{c o s (y)} - e^{s i n (x)} = sin (y)

What is the general solution for cos^2(x)=((4k-3))/8 ?
The general solution for cos^2(x)=((4k-3))/8 is